Answer:
Options (A), (B) and (C)
Explanation:
In an ecosystem, the transfer of nutrients takes place from one place o another. This input and output of nutrients continuously takes place in the ecosystem.
The rainwater is also essential as it carries a certain amount of valuable nutrients such as calcium, fluoride, magnesium, and iron that are mixed with water, and this enriches the fertility of the soil and the water is taken up by the roots of the plants for the growth and expansion of plants. The nitrogen-fixing bacteria refers to those micro-organisms that are present in the soil and helps in the conversion of the atmospheric nitrogen into nitrogen compounds, which are helpful for the enhancement of the soil fertility. Chemical weathering usually refers to the process by which the rocks disintegrates due to the effect of chemical reactions. The micro-organisms and the plant roots release some acids sometimes that react with the rocks and help in breakdown releasing the nutrients that are stored in the rocks. This eventually mixes up with the soil and increases the quality of the soil.Thus, the correct answers are options (A), (B) and (C).
Examples of nutrient input into a terrestrial ecosystem include nutrients dissolved in rain, nitrogen fixation by bacteria, and chemical weathering of rocks by organic acids released by plant roots.
Explanation:An example of nutrient input into a terrestrial ecosystem includes:
Nutrients dissolved in rain: Rain can carry dissolved nutrients such as nitrogen and phosphorus from the atmosphere into the soil, providing a source of nutrients for plants and other organisms.Nitrogen fixation by bacteria: Certain bacteria called nitrogen-fixing bacteria convert atmospheric nitrogen into a form that can be used by plants. This process provides plants with a source of nitrogen for growth.Chemical weathering of rocks by organic acids released by plant roots: Organic acids released by plant roots can break down rocks, releasing nutrients such as potassium and calcium into the soil.Learn more about Nutrient input into a terrestrial ecosystem here:https://brainly.com/question/31232676
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A piece of silver with a mass 437 g has a heat capacity of 103.6 J/°C. What is the specific heat of silver?
Answer:
Specific heat of silver is 0.237 J/g°C
Explanation:
To define the heat capacity (c) we have to know that it depends on the amount of mass.
If our system consists of a single substance, we talk about specific heat (C) from the relationship:
c = C. m
103.6 J/°C = C . 437 g
103.6 J/°C / 437 g = C
C = 0.237 J/g°C
Solution A is 0.44 M and reacts with 0.11 M of solution B. Assume that the value of x is 0, the value of y is 1, and r is 1.07 × 10⁻¹ /s. Find the rate constant for this reaction.
Final answer:
The rate constant for the reaction is 0.973 L/mol/s.
Explanation:
In the given question, we are provided with the initial concentration of solution A and solution B, as well as the values of x, y, and r. To find the rate constant for this reaction, we can use the rate equation:
rate = k[A]^x[B]^y
Substituting the given values:
0.107/s = k(0.44 M)^0(0.11 M)^1
Simplifying the equation:
0.107/s = k(0.11 M)
Dividing both sides by 0.11 M:
k = 0.107/s / 0.11 M = 0.973 L/mol/s
Lemon juice has a pH of about 2.0, compared with a pH of about 1.0 for stomach acid. Therefore, the concentration of H in stomach acid is __________ than that of lemon juice.
Answer: the concentration of H in the stomach acid is greater than that of the lemon juice
Explanation:Please see attachment for explanation
7. An example of a compound is _____. a) chicken noodle soup b) powerade c) air inside a balloon d) lead pipe e) baking soda (NaHCO3)
Answer:
E
Explanation:
A. Is wrong
To prepare chicken noodle soup, several things are needed to be mixed. This is what makes it a mixture
B is wrong
Powerade is not a compound.
C is wrong
The air inside a balloon is usually helium which is an element and not a compound
D. Lead pipe is not a compound
E. Baking soda is a compound as it contains elements in different ratios
Answer:baking soda
Explanation:
A compound is formed by chemical reaction of atoms of elements. A compound is actually formed by chemical combination of elements. A soul, a Powerade, air and a lead like are not compounds. In the first three items mentioned, on!y a mixture of substances are involved. There isn't any chemical combination at all. Lead pipe only consists of one kind of element. No other thing combines with it at all.
Biological oxidation-reduction reactions always involve:
a. direct participation of oxygen.
b. transfer of electron(s).
c. mitochondria.
d. transfer of hydrogens.
e. formation of water.
Answer:
b. transfer of electron(s).
Explanation:
An oxidation-reduction also called a redox reaction is a
chemical reaction in which electrons are transferred of between two species of reactants. It is a chemical reaction where the oxidation number of an atom, ion, or molecule, increases or decreases by losing or gaining electrons
Biological oxidation-reduction reactions always involve transfer of electrons.
Oxidation involves loss of electrons in which the matter then becomes
positively charged . On the other hand, reduction involves the gain of
electrons and the matter becomes negatively charged.
Oxidation-reduction(Redox) reactions is an important aspect of reactivity of
elements. Atoms begin to react when movement of electrons occurs which
gives rise to their loss and gain.
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If 18.1 g of ammonia is added to 27.2 g of oxygen gas, how many grams of excess reactant is remaining once the reaction has gone to completion?
Answer:
m of NH3 = 6.46 g
Explanation:
First, in order to know the limiting and excess reactant, we need to write and balance the equation that is taking place:
NH₃ + O₂ ---------> NO + H₂O
Now, let's balance the equation:
4NH₃ + 5O₂ ---------> 4NO + 6H₂O
Now that we have the balanced equation, let's see which reactant is in excess. To know that, let's calculate the moles of each reactant using the molar mass:
MM NH3 = 17 g/mol
MM O2 = 32 g/mol
moles NH3 = 18.1 / 17 = 1.06 moles
moles O2 = 27.2 / 32 = 0.85 moles
Now, let's compare these moles with the theorical moles that the balanced equation gave:
4 moles NH3 --------> 5 moles O2
1.06 moles ----------> X
X = 1.06 * 5 / 4 = 1.325 moles of O2
These means in order to NH3 completely reacts with O2, it needs 1.325 moles of O2, which we don't have it. We only have 0.85 moles of O2, therefore, the limiting reactant is the O2 and the excess is NH3.
Now, let's see how many grams in excess we have left after the reaction is complete.
4 moles NH3 --------> 5 moles O2
X moles NH3 ----------> 0.85 moles
X = 0.85 * 4 / 5 = 0.68 moles of NH3
This means that 0.85 moles of O2 will react with only 0.68 moles of NH3, and we have 1.06 so, the remaining moles are:
moles remaining of NH3 = 1.06 - 0.68 = 0.38 moles
Finally the mass:
m = 0.38 * 17
m = 6.46 g of NH3
At 1:00 pm I measure 10 grams of the element, but at 1:24 pm I measure only 1.25 grams of the element (with 8.75 grams of daughter product). How long is each half life, in minutes? (Just enter the number).
Answer:
The half life of the element = 8 minutes
Explanation:
Solution:
The initial amount = 10g
final amount = 1.25g
elapsed time = 24 minutes
Half life formula
t1/2 = t/(log1/2(Nt/N0))
or (Nt/N0) = 0.5^(t/(t1/2))
or 1.25/10 = 0.5^((24×60)/(t1/2))
Log of both sides gives
0.125 = 0.5^(1440/(t1/2))
-0.903 = 1440/t1/2×log(0.5)
-0.903 = 1440/t1/2×(-0.301)
3 = 1440/t1/2
t1/2 = 1440/3 = 480
Solving we have
t1/2 or the half life = 480s
= 480/60 minutes or
= 8 minutes
dvanced treatment of wastewater can include any of the following EXCEPT
1. reverse osmosis.
2. sand and gravel filtration.
3. phytoremediation.
4. charcoal filtration.
Answer:
2. sand and gravel filtration.
Explanation:
Sand and/or gravel filters are made at home usually comprising of a pipe or tub filled with sand and gravel, while the advanced commercial systems usually filters contain other filtration media and processes such as charcoal filtration phytoremediation reverse osmosis
Wet oxidation
Activated sludge systems
Upflow anaerobic sludge
Vacuum evaporation
blanket digestion
Anaerobic filter
Urine-diverting dry toilet
Vermifilter
Activated sludge model
This triple beam balance is used to measure ________ in grams. A) density B) mass C) volume D) weight
Answer:
B) mass
Explanation:
Triple beam balance -
It is an instrument , which is used for the measurement of the mass , in a very accurate and precise manner .
The error range of this balance is +/- 0.05 gram .
It consists of three beam , of three different sizes , where the difference in the beam shows the difference in the weights and readings .
Hence, from the given statement of the question,
The correct option is B. mass.
This triple beam balance is used to measure mass in grams. Option B.
Functions of the triple beam balanceA triple beam balance is a common laboratory instrument used to measure the mass of an object. Mass is a fundamental property of matter and represents the amount of substance in an object.
The triple beam balance consists of three beams, each with different sliding weights. By adjusting these weights, the balance allows for the measurement of an object's mass in grams.
Mass is distinct from weight, which is the force exerted on an object due to gravity and can vary depending on the gravitational pull.
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Ethylene (C2H4) burns in oxygen to produce carbon dioxide and water. What is the correct form of the balanced chemical equation that describes this reaction?
Answer:
The answer to your question is C₂H₄ + 3O₂ ⇒ 2CO₂ + 2H₂O
Explanation:
Data
Ethylene = C₂H₄
Oxygen = O₂
Carbon dioxide = CO₂
Water = H₂O
Reaction
C₂H₄ + O₂ ⇒ CO₂ + H₂O
Reactant Element Product
2 Carbon 1
4 Hydrogen 2
2 Oxygen 3
This reaction is unbalanced
C₂H₄ + 3O₂ ⇒ 2CO₂ + 2H₂O
Reactant Element Product
2 Carbon 2
4 Hydrogen 4
2 Oxygen 6
Now, the reaction is balanced
To save time you can approximate the initial volume of water to ±1 mL and the initial mass of the solid to ±1 g. For example, if you are asked to add 23 mL of water, add between 22 mL and 24 mL. Which metals in each of the following sets will have equal density?
1. 20.2 g gold placed in 21.6 mL of water and 12.0 g copper placed in 21.6 mL of water.
2. 20.2 g silver placed in 21.6 mL of water and 12.0 g silver placed in 21.6 mL of water.
3. 15.2 g copper placed in 21.6 mL of water and 50.0 g copper placed in 23.4 mL of water.
4. 15.4 g gold placed in 20.0 mL of water and 15.7 g silver placed in 20.0 mL of water.
5. 20.2 g silver placed in 21.6 mL of water and 20.2 g copper placed in 21.6 mL of water.
6. 11.2 g gold placed in 21.6 mL of water and 14.9 g gold placed in 23.4 mL of water.
Answer:
The correct answers are option 2, 3 and 6.
Explanation:
Density is defined as mas of substance present in an unit volume of the substance.
[tex]Density=\frac{Mass}{Volume}[/tex]
Density of same substance with different masses and volume remains the same that is it is an intensive property.
2. 20.2 g silver placed in 21.6 mL of water and 12.0 g silver placed in 21.6 mL of water.
3. 15.2 g copper placed in 21.6 mL of water and 50.0 g copper placed in 23.4 mL of water.
6. 11.2 g gold placed in 21.6 mL of water and 14.9 g gold placed in 23.4 mL of water.
Since the metal kept in both the cases are same.And metal has fix value of density , so from the given options the the option with sets of same of metals has equal densities.
if 1.72 mol of ZnS is heated in the presence of 3.04 mol pf O2,which is the limiting reactant?
Answer:
The limiting reactant is the ZnS
Explanation:
The equation for this reaction is:
2 ZnS + 3O₂ = 2 ZnO + 2 SO₂
2 moles of zinc sulfure reacts with 3 moles of oxygen.
Then, 1.72 mol of ZnS would react with ( 1.72 .3)/2 = 2.58 moles of O₂
If we have 3.04 moles, then the oxygen is the reactant in excess.
Let's confirm, the ZnS as the limiting reactant.
3 moles of oxygen react with 2 moles of sulfure.
Then, 3.04 moles of O₂ would react with (3.04 .2) / 3 = 2.02 moles of ZnS
We have 1.72 moles of Zn S and it is not enough for the 2.02 moles that we need, for the reaction.
If a hazardous bottle is labeled 20.2 percent by mass Hydrochloric Acid, HCl, with a density of 1.096 g/mL, calculate the molarity of the HCl solution.
Answer:6M
Explanation:
From Co= 10pd/M
Where Co= molar concentration of raw acid
p= percentage by mass of raw acid=20%
d= density of acid=1.096g/cm3
M= molar mass of acid=36.5
Co= 10×20×1.096/36.5=6M
To calculate the molarity of the HCl solution, convert the given mass percentage to grams, then convert grams to moles using the molar mass of HCl, finally divide the moles of HCl by the volume of the solution in liters to obtain the molarity. In this case, the molarity of the HCl solution is approximately 0.576 M.
Explanation:To calculate the molarity of the HCl solution, we need to convert the given mass percentage to grams. If we assume we have 100 grams of the solution, then 20.2 grams would be HCl. Next, we convert grams to moles using the molar mass of HCl (36.46 g/mol). Finally, we divide the moles of HCl by the volume of the solution in liters to obtain the molarity.
Molarity (M) = (moles of HCl) / (volume of solution in liters)
In this case, the molarity of the HCl solution would be calculated as:
Molarity = (20.2 g / 36.46 g/mol) / (1 L * 1.096 g/mL)
After simplifying the expression, the molarity of the HCl solution is approximately 0.576 M.
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A block weighing 8.5 N requires a force of 2 Nto push it along at constant velocity.What is the coefficient of friction for thesurface?
Answer:
The coefficient of friction of the surface μ = 0.235
Explanation:
Values of the given variables:
Weight of block = 8.5 N
Pushing force F = 2 N
velocity = constant
Solution:
Newton's Second Law of motion states that acceleration of an object depends on the mass of the object and the force applied
F = m a
Since v =constant, a = 0. Thus
Net force Fnet = 0.
Now F_net is the sum of the pushing force and the frictional force along the horizontal, which opposes the motion:
Fnet = Fpushing - Ffrictional_force
Given that Fnet = 0, Fpushing = Ffrictional_force. and Ffrictional_force = 2 N
Equation of friction is given by
Ffrictional_force = μ*W
2 = μ*8.5
μ = 2/8.5 = 0.235
The coefficient of friction for the surface is approximately 0.235.
Explanation:The coefficient of friction can be determined using the equation:
frictional force = coefficient of friction × normal forceGiven that the weight of the block is 8.5 N, and the force required to push it at constant velocity is 2 N, we can determine the normal force as follows:
normal force = weight of the blocknormal force = 8.5 NSubstituting the values into the equation:
2 N = coefficient of friction × 8.5 NSolving for the coefficient of friction:
coefficient of friction = 2 N / 8.5 Ncoefficient of friction ≈ 0.235Therefore, The coefficient of friction for the surface is approximately 0.235.
The student measures the actual pressure of the CO2(g) in the container at 425 K and observes that it is less than the pressure predicted by the ideal gas law. Explain this observation.
Real gases do not obey the ideal gas law
The pressure is less than predicted because of the attractive forces between the molecules as well as energy is lost during collision of a real gas
Reason:
The ideal gas equation is given as follows;
P·V = n·R·TThe real gas equation is presented as follows;
[tex]\left[P + \dfrac{a \cdot n^2}{V^2} \right]\cdot \left[V - n \cdot b\right] = n \cdot R \cdot T[/tex]The coefficient +a, is the pressure correction factor, given that the pressure of a real gas is lower than that of an ideal, due to the attractive force that exist between the gas molecules that reduces the number of collisions of the molecules with the container's wall when they attract each other
The collisions that occur in real gas involve energy loss, therefore, the molecules tend to be less activated as predicted by the real gas law
Therefore;
The pressure of the real CO₂ gas is expected to be much lesser than expected and has to be corrected using the correction factor for the value correspond with the predicted value at the given temperatureLearn more here:
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Write the net ionic equation for this reaction occurring in water: Sodium phosphate and calcium chloride are mixed to form sodium chloride and calcium phosphate.
1. No reaction occurs.
2. 2Na3PO4 + 3Ca2+ -> 6Na+ + Ca3(PO4)2
3. 2 PO3−4 + 3Ca2+ -> Ca3(PO4)2
4. PO2−4 + Ca2+ -> CaPO4
5. 2 PO3−4 + 3CaCl2 -> 6Cl− + Ca3(PO4)2
Final Answer:
The net ionic equation for the reaction between sodium phosphate and calcium chloride is 2 PO3−4 + 3Ca2+ → Ca3(PO4)2
Explanation:
The net ionic equation for the reaction between sodium phosphate and calcium chloride in water is:
2 PO3−4 + 3Ca2+ -> Ca3(PO4)2
This equation represents the formation of solid calcium phosphate and the dissociation of the phosphate ions. The net ionic equation showcases the essential chemical transformation by emphasizing the interaction between phosphate ions and calcium ions, resulting in the precipitation of calcium phosphate. It succinctly highlights the key species undergoing change in the reaction, streamlining the representation of the chemical process. Spectator ions, such as sodium and chloride, are excluded from the equation as they remain unchanged during the reaction and do not contribute to the formation of the precipitate.
The correct answer is option 3. The net ionic equation for the reaction occurring in water is [tex]\[ 2 \text{PO}_4^{3-} + 3 \text{Ca}^{2+} \rightarrow \text{Ca}_3(\text{PO}_4)_2 \][/tex].
To determine the net ionic equation, we need to consider the solubility of the compounds in water and the formation of precipitates. Sodium phosphate [tex](Na_3PO_4)[/tex] and calcium chloride [tex](CaCl_2)[/tex] are both ionic compounds and dissociate into their constituent ions in water.
Sodium phosphate dissociates into sodium ions [tex](Na+)[/tex] and phosphate ions [tex](PO4^3-)[/tex]. Calcium chloride dissociates into calcium ions [tex](Ca^2+)[/tex] and chloride ions [tex](Cl^{-})[/tex].
When these solutions are mixed, the calcium ions from calcium chloride react with the phosphate ions from sodium phosphate to form calcium phosphate [tex](Ca_3(PO_4)_2)[/tex], which is a precipitate. Sodium chloride (NaCl) remains dissolved in the solution because it is highly soluble in water.
The overall molecular equation for the reaction is:
[tex]\[ 2 \text{Na}_3\text{PO}_4 + 3 \text{CaCl}_2 \rightarrow 6 \text{NaCl} + \text{Ca}_3(\text{PO}_4)_2 \][/tex]
By removing the spectator ions, we are left with the net ionic equation that shows only the ions that actually participate in the formation of the precipitate:
[tex]\[ 2 \text{PO}_4^{3-} + 3 \text{Ca}^{2+} \rightarrow \text{Ca}_3(\text{PO}_4)_2 \][/tex]
Manganese nodules are considered a valuable resource from the ocean floor because they are abundant and contain not only manganese but also?
Answer: iron (oxides of iron)
Explanation: manganese nodules usually contain layers of iron and manganese oxides in a concentric arrangement and are called polymetallic nodules.
A student obtains a mixture of the chlorides of two unknown metals, X and Z. The percent by mass of X and the percent by mass of Z in the mixture is known. Which of the following additional information is most helpful in calculating the mole percent of XCl(s) and of ZCl(s) in the mixture?
Answer:
the Molar masses of Z and X
Explanation:
The percent by mass of X states how many grams of X are in 100 grams of the mixture. Then, we know how many grams of X, Z (we also know its percent) and of Cl ( = 100 - X % - Z %). Taking 100 g of the mixture as a base, we have:
x grams of X
z grams of Z
y grams of Cl
With the molar masses of Z and X (and of Cl) we can transform all masses into moles and then compute the mole percent of XCl(s) and of ZCl(s) in the mixture
From the given information:
The percent by mass of X = KnownThe percent by mass of Z = KnownThe percentage mass is given as the mass of the element which is divided by mass of compound multiplied by 100.
it is estimated using the formular:
[tex]\mathbf{percentage \ mass = \dfrac{\text{mass of element in 1 mole of compound}} { \text{mass of 1 mole of compound}} \times 100}[/tex]
From there;
The mole percentage is determined, which is the ratio of moles in a substance (element or compound) divided by the total amount of the mixture.
Using the formula:
Mole percentage = mole fraction × 100
Since the mole fraction is not given, it can be estimated as well by using the formula:
[tex]\mathbf{Mole\ fraction = \dfrac{\text{No of moles of a component} }{\text{total number of moles of the compound}}}[/tex]
Eventually; To determine the unknown number of moles, we will need the value of their respective molar masses(i.e. for X and Z) by using the formula:
[tex]\mathbf{Number \ of \ moles = \dfrac{Mass}{Molar \ Mass}}[/tex]
In conclusion, the additional information required in calculating the mole percent of XCl(s) and ZCl(s) in the mixture is the molar mass of X and Z.
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Broad-spectrum pesticides: kill beneficial organisms in addition to the targeted pest. are readily broken down into safe materials such as water and CO2. are ineffective.
Answer:
kill beneficial organisms in addition to the targeted pest
Explanation:
The use of pesticides have emerged as one of the main options used in the control of invertebrate pests in different parts of the world. Organophosphate is the mostly used form of pesticide in places like Australia. There have been reported issues of crops losses from the incidences of pests, despite the increase in the use of pesticides. Broad spectrum pesticides end up kill non-target invertebrate species together with the insects and mites they are out to eliminate
The removal of silver tarnish from silverware using aluminum foil and a solution of electrolyte is an example of
A) oxidation of aluminum metal.
B) reduction of silver ions.
C) an electrochemical cell.
D) all of the above
Answer:
The answer is C. An electrochemical cell.
Explanation:
The aluminum ion react with the sulfide to form aluminum sulfide.
Calculate the vapor pressure of water above a solution prepared by adding 24.5 g of lactose (C12H22O11) to 200.0 g of water at 338 K. (Vapor-pressure of water at 338 K 187.5 torr.)
Answer:
Vapor pressure of solution → 186.3 Torr
Explanation:
To solve this problem we must apply the colligative property of vapor pressure.
ΔP = P° . Xm
Where ΔP = Vapor pressure of pure solvent - Vapor pressure of solution
P° is vapor pressure of pure solvent → 187.5 Torr
Xm is the mole fraction of solute (moles of solute / total moles)
Let's determine the mole fraction,
Moles of solute = Mass of solute / Molar mass
24.5 g / 342 g/mol = 0.0716 moles
Moles of solvent = Moles of solvent /Molar mass
200 g / 18g/mol = 11.1 moles
Total moles = 11.1 moles + 0.0716 moles → 11.1716 moles
Mole fraction of solute = 0.0716 mol / 11.1716 mol = 6.40×10⁻³
Let's apply the formula
ΔP = P° . Xm
Vapor P of pure solvent - Vapor P of solution = P° . 6.40×10⁻³
Vapor P of pure solvent - 187.5 Torr . 6.40×10⁻³ = Vapor P of solution
187.5 Torr - 187.5 Torr . 6.40×10⁻³ = Vapor P of solution
Vapor pressure of solution → 186.3 Torr
The vapor pressure has been the pressure exerted by the molecules in the solution in vapor phase. The vapor pressure of water in solution is 186.3 torr
What are colligative properties?The colligative properties of the solution are dependent on the solute particles in the solution. The colligative properties include boiling point, freezing point, vapor pressure and osmotic pressure.
The change in the vapor pressure of the solution with the addition of solute ([tex]\Delta P[/tex]) is given as :
[tex]\Delta P=P^\circ\;\times\;x_m[/tex]
Where, the vapor pressure of the pure solvent, [tex]P^\circ=187.5\;\rm torr[/tex]
The mole fraction of the solute ([tex]x_m[/tex]) is given as:
[tex]x_m=\rm \dfrac{Moles\;solute}{Total\;moles}[/tex]
The moles of lactose in 24.5 grams of sample is:
[tex]\rm Moles\;solute=\dfrac{mass}{molar\;mass}\\\\ Moles\;lactose=\dfrac{24.5}{342\;g/mol} \\\\Moles\;Lactose=0.0706\;moles[/tex]
The moles of solvent water is given as:
[tex]\rm Moles\;solvent\;(water)=\dfrac{200}{18\;g/mol} \\Moles\;solvent=11.1\;mol[/tex]
The total moles of the solution is given as:
[tex]\rm Total \;moles=solute+solvent\\Total\;moles=0.0706+11.1\;mol\\Total \;moles=11.1706\;mol[/tex]
The moles fraction of solute is given as:
[tex]x_m=\dfrac{0.0706}{11.1706}\\\\ x_m=6.40\;\times\;10^{-3}[/tex]
The vapor pressure change in the solution is calculated as:
[tex]\Delta P=187.5\;\times\;6.40\;\times\;10^-^3\\\Delta P=1.2\;\rm torr[/tex]
The vapor pressure of the solution is given as:
[tex]\Delta P=\rm Solution\;vapor\;pressure-solvent\;vapor\;pressure\\1.2=Solution\;vapor\;pressure-187.5\;torr\\Solution\;vapor\;pressure=186.3\;torr[/tex]
The vapor pressure of the solution is 186.3 torr.
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Hydrocarbons, compounds containing only carbon and hydrogen, are important in fuels. The heat of combustion of cyclopropane, C3H6, is 499.8 kcal/mol. Write a balanced equation for the complete combustion of cyclopropane. How much energy is released during the complete combustion of 497 grams of cyclopropane? kcal
Answer:
The complete combustion of cyclopropane is:
2C₃H₆ + 9O₂ → 6CO₂ + 6H₂O
5914.3kcal are realeased in the combustion of 497 g of cyclopropane
Explanation:
Heat of combustion = 499.8 kcal/mol
This data means that 1 mol of C₃H₆ release 499.8 kcal.
Let's convert the mass to moles (mass / molar mass)
497 g / 42g/mol = 11.83 moles
Let's make a rule of three:
1 mol ____ 499.8
11.83 moles _____ (11.83 .499.8)/1 =5914.3kcal
The balanced equation for the complete combustion of cyclopropane:
C3H6 + 5O2 → 3CO2 + 3H2O
5748.7 kcal of energy is released during the complete combustion of 497 grams of cyclopropane.
The heat of combustion of cyclopropane is -499.8 kcal/mol, which means that 499.8 kcal of heat is released when 1 mole of cyclopropane is combusted.
To determine how much energy is released during the complete combustion of 497 grams of cyclopropane, we first need to determine the number of moles of cyclopropane in 497 grams.
The molar mass of cyclopropane is 42.08 grams/mol, so 497 grams of cyclopropane is equal to 497 / 42.08 = 11.5 moles of cyclopropane.
The energy released = (number of moles of cyclopropane) * (heat of combustion per mole of cyclopropane)
= 11.5 moles * 499.8 kcal/mol
= 5748.7 kcal
Therefore, 5748.7 kcal of energy is released during the complete combustion of 497 grams of cyclopropane.
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If 0.255 moles of AgNO₃ react with 0.155 moles of H₂SO₄ according to this UNBALANCED equation below, how many grams of Ag₂SO₄ could be formed? AgNO₃(aq) + H₂SO₄ (aq) → Ag₂SO₄ (s) + HNO₃ (aq)
After balancing the chemical equation, we determine the limiting reactant, which is H₂SO₄ in this case. This produces 0.155 moles of Ag₂SO₄. When converted to mass, this is approximately 48.33g of Ag₂SO₄.
Explanation:To answer this question, we first need to balance the chemical equation. The balanced equation is: 2 AgNO₃(aq) + H₂SO₄ (aq) → Ag₂SO₄ (s) + 2 HNO₃ (aq).
From the balanced equation, we can see that 2 moles of AgNO₃ react with 1 mole of H₂SO₄ to produce 1 mole of Ag₂SO₄. Thus, the mole ratio of AgNO₃ : H₂SO₄ : Ag₂SO₄ is 2:1:1.
Given that we have 0.255 moles of AgNO₃ and 0.155 moles of H₂SO₄, it’s clear that H₂SO₄ is the limiting reactant as it produces fewer moles of Ag₂SO₄ (based on the stoichiometric ratios). Therefore, 0.155 moles of Ag₂SO₄ will be formed.
To convert this to grams, we multiply by the molar mass of Ag₂SO₄, which is 311.8 g/mol: 0.155 moles * 311.8 g/mol = 48.33 grams of Ag₂SO₄ can be formed.
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To calculate the grams of Ag2SO4 formed, we first determine the limiting reactant from the balanced equation and then convert the moles of the limiting reactant to grams using its molar mass, resulting in 48.329 grams of Ag2SO4.
The question asks how many grams of Ag2SO4 could be formed when 0.255 moles of AgNO3 reacts with 0.155 moles of H2SO4. To solve this, we must balance the chemical equation and determine the limiting reactant.
First, we balance the equation:
2 AgNO3(aq) + H2SO4 (aq) → Ag2SO4 (s) + 2 HNO3 (aq)
With this balanced equation, we see that two moles of AgNO3 are needed to react with one mole of H2SO4. Since we have 0.255 moles of AgNO3 and 0.155 moles of H2SO4, H2SO4 is the limiting reactant. The mole ratio of H2SO4 to Ag2SO4 is 1:1, therefore 0.155 moles of Ag2SO4 will be formed.
To find the mass of Ag2SO4 formed, we use its molar mass (311.8 g/mol):
0.155 moles × 311.8 g/mol = 48.329 grams of Ag2SO4.
Mendeleev organized the elements
a. alphabetically by name.
b. by the number of electrons.
c. by increasing the atomic number and similar properties.
d. by increasing atomic weight and similar properties.
Answer:
D
Explanation:
Mendeleev was a Russian scientist. His works provided the framework for the present day periodic table. What Mendeleev did was to arrange the elements in order of their atomic weights.
He however had some gaps in his periodic table. He impressively predicted the elements that would fill in these blank spaces even though they were yet to be discovered as at then.
Give the outer electron configuration for each of the following columns in the periodic table. 1A,2A,5A,7A Express your answer as a string without blank space between orbitals. For example, the outer electron configuration for the column 4A, ns2np2, should be entered as ns^2np^2.
Answer:
1A: ns^1
2A: ns^2
5A: ns^2np^3
7A: ns^2np^5
Explanation:
According to IUPAC, the columns 1A, 2A, 5A, and 7A correspond to the groups, 1, 2, 15 and 17, respectively.
The outer electron configuration of these columns are the following:
Column Outer electron configuration
1A ns^1
2A ns^2
5A ns^2np^3
7A ns^2np^5
Therefore, for the column 1A the s orbital has only one electron, for the column 2A has the s orbital completed with 2 electrons, for the column 5A the number of electrons in the s orbital is complete (2) and the number of electrons in the p orbital is 3, for the column 7A the s orbital has 2 electrons and the p orbital has 5 electrons.
I hope it helps you!
The outer electron configurations for columns 1A, 2A, 5A, and 7A in the periodic table are ns¹, ns², ns²np³, and ns²npµ respectively, where 'n' corresponds to the period number.
Explanation:The outer electron configuration for the columns in the periodic table you've asked about can be predicted based on their position.
For column 1A (alkali metals), the outer configuration is ns¹.For column 2A (alkaline earth metals), the outer configuration is ns².For column 5A (pnictogens), the outer configuration is ns²np³.For column 7A (halogens), the outer configuration is ns²npµ.These configurations show the distribution of electrons in the outermost shell of atoms in each respective column. Remember that 'n' represents the quantum level corresponding to the period number of the elements in the column.
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Check all that apply.
a. Cathode rays have mass.
b. There are subatomic size particles in an atom.
c. In the atom, the particles of the cathode rays are embedded in a diffuse cloud of positive charge.
d. Matter contains negative charge and by inference positive charge.
e. Particles of the cathode rays are fundamental to all matter.
Answer:
a.Cathode rays have mass.
b.They are subatomic size particles in an atom.
d. Matter contains negative charge and by inference positive charge.
e. Particles of the cathode rays are fundamental to all matter.
Explanation:
The statements above reflect the Thompson model of the atom. Electrons were first known as cathode rays. The Thompson's model viewed the atom as a sphere of negative charges into which positive charges were embedded. This is the plum pudding model of the atom.
____ are weak bonds that are not strong enough to hold atoms together to form molecules but are strong enough to form bridges between molecules. a. Hydrogen bonds
b. Covalent bonds
c. Ionic bonds
d. Anionic bonds
e. Polar covalent bonds
Answer:Hydrogen bonds
Explanation:
Hydrogen bonds occur between molecules when hydrogen is covalently bonded to a highly electronegative element in each molecule of the compound E.g water. Hydrogen bonds are intermolecular bond forces. They hold molecules of a substance together in a particular state of matter. They do not link atoms together but they form bridges between molecules. Hydrogen bonds exist in liquid, solid and gaseous states of matter.
Hydrogen bonds are weak bonds that form bridges between molecules. They occur when a hydrogen atom is attracted to an atom with a partial negative charge.
Explanation:Hydrogen bonds are weak bonds that are not strong enough to hold atoms together to form molecules but are strong enough to form bridges between molecules. These bonds occur when a hydrogen atom with a partial positive charge is attracted to an atom with a partial negative charge.
An example of hydrogen bonding is the bond between water molecules. The partial positive charge of the hydrogen atom in one water molecule is attracted to the partial negative charge of the oxygen atom in a neighboring water molecule, resulting in hydrogen bonding.
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You want to create the effect of the sun shining on the actors in a theater production. What lighting would you use to create this effect?
Answer:
Metal Halide arc lamp
Explanation:
Metal Halide arc lamp:
Metal Halide arc lamps are made for film and television lighting production use in where daylight colour match and a high temporal stability are required, they are also used in solar simulation for testing sunscreens, plastics, solar cells as well as other devices and production materials
A 44.0mg sample of carbon reacts with sulfur to form 122 mg of the compound.
What is the empirical formula of the carbon sulfide? Express answer as a chemical formula.
Answer:
Empirical formula = C₃S₂
Explanation:
Given data:
Mass of carbon = 44.0 mg (44/1000 = 0.044 g)
Mass of sulfur = 122 mg - 44.0 mg = 78 mg = 78/1000 = 0.078 g)
Empirical formula = ?
Solution:
First of all we will calculate the number of moles.
Number of moles of carbon = mass / molar mass
Number of moles of carbon = 0.044 g/ 12.01 g/mol
Number of moles of carbon = 0.0037 mol
Number of moles of sulfur:
Number of moles = mass / molar mass
Number of moles = 0.078 g/ 32,066 g/mol
Number of moles = 0.0024 mol
Now we will compare the moles:
C : S
0.0037/0.0024 : 0.0024/0.0024
1.5 : 1
C : S = 2(1.5 : 1)
C : S = 3 : 2
Empirical formula = C₃S₂
If a proton and an electron are released when they are 5.50×10−10 mm apart (typical atomic distances), find the initial acceleration of each of them.
Answer: The initial acceleration of the proton = (4.56 × 10^23) m/s2
The initial acceleration of the electron = (8.36 × 10^26) m/s2
Explanation: The force of attraction between the proton and electron can be computed using the statements of Coulomb's law which state that the force of attraction between two charged particles is directly proportional to the product of the two charges and inversely proportional to the square of their distances apart.
F = (Kq1q2)/(r^2) where K = (9 × (10^9) Nm(C^-2))
But q1 is the charge on a proton = (1.6 × (10^-19)) C
q2 is charge on an electron = -(1.6 × (10^-19)) C
r = (5.50 × (10^-10))mm = (5.50 × (10^-13))m
Computing all that, F = 0.0007616529 N = (7.62 × 10^-4) N
But the force of attraction is converted to that required for motion when they're released.
F = ma.
For proton, m = (1.67 × 10^-27) kg
a = F/m = 0.000762/(1.67 × 10^-27) = (4.56 × 10^23) m/s2
For electron, m = (9.11 × 10^-31) kg
a = F/m = 0.000762/(9.11 × 10^-31) = (8.36 × 10^26) m/s2
QED!
Final answer:
The initial acceleration of a proton and an electron, when released 5.50×10−10 mm apart, can be found using Coulomb's law and Newton's second law. The electrostatic force is first calculated using the charge of the particles and the separation distance, then divided by their respective masses to find their accelerations.
Explanation:
To calculate the initial acceleration of a proton and an electron when they are 5.50×10−10 mm apart, we use Coulomb's law which states that the electrostatic force (F) between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them. The formula is F = k × (q1 × q2) / r2, where k is the electrostatic constant (8.99×109 N×m2/C2), q1 and q2 are the charges of the proton and electron, and r is the separation distance. The acceleration (a) of each particle can then be found using Newton's second law (F = m × a), by dividing the electrostatic force by the mass (m) of the proton or electron respectively.
For a proton, the charge is +1.602×10−19 C and mass is 1.67×10−27 kg. For an electron, the charge is −1.602×10−19 C and mass is 9.11×10−31 kg. The initial acceleration for each particle can be calculated by using their respective masses and the electrostatic force derived from Coulomb's law.