Answer:
D. Ne
Explanation:
Ne Across a period, increasing nuclear charge outweighs the shielding. So the outermost electrons are held more tightly and ionization energy increases as we go across a period. Ne, being placed at the extreme right end of the second period, has the highest ionization energy among the provided options.
When the mineral dolomite (caco3·mgco3) is heated, it gives off carbon dioxide and forms a mixture of a metal oxide and a metal carbonate. which carbonate is formed?
How much energy is needed to melt 25.4 grams of I2?
Hfus=61.7 J/g
Type the correct answer to 3 digits. If the answer has an exponent, type the number, then ^, then the exponent.
(The picture might make the question more clearer.)
The amount of energy need to melt 25.4 grams of iodine, I₂ is 1.57×10³ Joules
How to calculate the amount of energy needed to to melt the the iodine, I₂?
The following useful information were given from the question:
Mass of iodine, I₂ (m) = 25.4 gramsLatent heat of fusion (Hf) = 61.7 J/gHeat energy needed to melt the iodine, I₂ (Q) =?The amount of energy need to melt 25.4 grams of iodine, I₂ can be calculated as shown below:
Heat energy needed (Q) = Mass (m) × Latent heat of fusion (Hf)
= 25.4 grams × 61.7 J/g
= 1.57×10³ Joules
Thus, the amount of heat energy needed to melt the iodine,I₂ is 1.57×10³ Joules
The reaction of 1.250 g k with o2 forms 2.273 g of an oxide of potassium that is used in self-contained breathing devices. determine the formula for this oxide of potassium.
Engineers often attempt to capture and reuse energy or to use renewable resources to minimize harm to the environment. This is referred to as
How many moles of aspartame are present in 5.00 mg of aspartame?
To find the number of moles in a given mass, convert the mass to grams (if necessary) and divide by the molar mass. Hence, in 5.00 mg of aspartame, there are approximately 1.70 x 10^-5 moles.
Explanation:To calculate the number of moles of aspartame in a given mass, we would use the molar mass of aspartame. First, we need to convert the mass from milligrams to grams because molar mass is usually given in g/mol. As 1 gram (g) is equal to 1000 milligrams (mg), 5.00 mg is equal to 0.005 g.
The molar mass of aspartame (C14H18N2O5) is approximately 294.31 g/mol. After converting the mass to grams, we use the formula: number of moles = mass (g) / molar mass (g/mol).
Therefore, the number of moles of aspartame in 5.00 mg is 0.005g / 294.31 g/mol = 1.70 x 10^-5 mol. So, there are roughly 1.70 x 10^-5 moles of aspartame in 5.00 mg.
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Calculate how many grams of o2(g) can be produced from heating 87.4 grams of kclo3(s).
Final answer:
By applying stoichiometry and molar mass conversions, it is determined that heating 87.4 grams of KClO₃(s) will produce 34.24 grams of O₂(g).
Explanation:
Calculating Grams of Oxygen Gas from Potassium Chlorate Decomposition
To determine how many grams of O₂(g) can be produced by heating 87.4 grams of KClO₃(s), we need to understand the stoichiometry of the chemical reaction involved. The balanced chemical equation for the decomposition of potassium chlorate (KClO₃) is:
2 KClO₃(s) → 2 KCl(s) + 3 O₂(g)
Using the molar masses: 1 mol KClO₃ = 122.55 g/mol and 1 mol O₂ = 32.00 g/mol, we can calculate the amount of oxygen gas produced from the given mass of potassium chlorate.
Convert the mass of KClO₃ to moles:
(87.4 g KClO₃) × (1 mol KClO₃ / 122.55 g KClO₃) = 0.713 moles of KClO₃
Use the stoichiometry of the reaction to find moles of O₂ produced:
(0.713 moles KClO₃) × (3 moles O₂ / 2 moles KClO₃) = 1.070 moles of O₂
Convert moles of O₂ to grams:
(1.070 moles O₂) × (32.00 g/mol O₂) = 34.24 grams of O₂
Hence, 34.24 grams of O₂ can be produced from heating 87.4 grams of KClO₃(s).
Which title of a poem is handwritten correctly?
a. "Afternoon on a Hill"
b. Afternoon on a Hill
c. Afternoon on a Hill
The correct way to write the title of a poem is by enclosing it in quotation marks, making option (a) "Afternoon on a Hill" the correct formatting for a handwritten poem title.
When formatting the title of a poem in writing, it should either be enclosed in quotation marks or italicized if typed, depending on the style guide being used. The three options given for the poem title 'Afternoon on a Hill' vary in formatting. Based on standard English conventions for titling creative works such as poems, the handwritten version of the poem title would be formatted using either quotation marks or italicization.
Therefore, the correct answer is option (a) "Afternoon on a Hill" as it follows the guideline of using quotation marks for the title of shorter works such as poems. Option (b) and option (c) are missing the necessary punctuation or formatting emphases that set apart the titles of creative works from the rest of the text.
What is the volume of 12.8 g of a liquid that has a density of 0.789 g/mL?
Answer: The volume of liquid is 16.22 mL.
Explanation:
Density is defined as the ratio of mass of the object and volume of the object.
Mathematically,
[tex]\text{Density}=\frac{\text{Mass of the object}}{\text{Volume of the object}}[/tex]
We are given:
Density of liquid = [tex]0.789g/mL[/tex]
Mass of liquid = 12.8 g
Putting values in above equation, we get:
[tex]0.789g/mL=\frac{12.8g}{\text{Volume of liquid}}\\\\\text{Volume of liquid}=16.22mL[/tex]
Hence, the volume of liquid is 16.22 mL.
Describe the relationship between pure chemistry and applied chemistry
Pure chemistry involves the study of the basic principles and theories of chemistry while applied chemistry uses these principles to solve real-world problems or develop new products. They are interconnected in the sense that advancements in pure chemistry can lead to new opportunities in applied chemistry and vice versa.
Explanation:The relationship between pure chemistry and applied chemistry is similar to that of theory and practice. Pure chemistry, also known as basic chemistry, is the study of the basic principles and theories of chemistry for the sake of knowledge itself. It seeks to understand the fundamental principles that explain how matter behaves and interacts. Examples include studying chemical reactions, the properties of elements, and how atoms form compounds.
On the other hand, applied chemistry utilizes these principles to solve real-world problems or develop new products. For instance, an applied chemist might work on creating new material, developing new drugs, improving environmental sustainability, or advancing food science.
In essence, applied chemistry is an application of pure chemistry. Both fields are interconnected; advancements in pure chemistry can lead to new opportunities in applied chemistry and vice versa.
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1 kg is equal to...
Options:
1) 103 hg.
2) 103 g.
3) 10-2 cg.
4) 10-2 g.
Use unit conversion to calculate the number of minutes in one week.
There are 10,080 minutes in one week. To calculate the number of minutes in one week, multiply the number of minutes in an hour (60) by the number of hours in a day (24), and then multiply that result by the number of days in a week (7).
Explanation:To calculate the number of minutes in one week using unit conversion, you first need to know how many minutes are in an hour, how many hours are in a day, and how many days are in a week. There are 60 minutes in an hour. In each day, there are 24 hours. Therefore, in one day, there are 60 x 24 = 1440 minutes. A week consists of 7 days, so to find out the number of minutes in a week, you would do 1440 x 7 = 10,080. So, there are 10,080 minutes in one week.
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These reactions can also be reversed, breaking the large molecules into its individual molecules. what substance would need to be added in order to reverse the reaction
Which subatomic particle is the identity of the atom and why?
An unknown molecule contains only carbon and sulfur, and it is 15.77% carbon by mass. what is the molecular geometry of this molecule?
We take as a basis for solving this item, 100 g of sample. If 15.77% of the sample is carbon, we solve for the amount of carbon in the sample,
Carbon = (100 g)(0.1577) = 15.77
Then, we solve for the amount sulphur in the sample by,
Sulfur = (100 g)(1 – 0.1577) = 84.23 g
Next, we solve for the number of moles of the elements by dividing the calculated masses by the molar mass.
Carbon = (15.77g)(1 mol/12 g) = 1.3 moles
Sulphur = (84.23 g)(1 mol/32 g) = 2.63 moles
From the ratio of the calculated number of moles, we can say that each mol of carbon will have to be paired up with two other sulphur atoms. Thus, the shape of the molecule is LINEAR.
Could you separate the sodium from the chlorine by crushing the salt crystals
Classify each of these soluble solutes as a strong electrolyte, a weak electrolyte, or a nonelectrolyte. solutes formula nitric acid hno3 lithium hydroxide lioh hydrofluoric acid hf ammonia nh3 potassium chloride kcl ethanol c2h5oh glucose c6h12o6
Final answer:
Nitric acid and lithium hydroxide are classified as strong electrolytes, meaning they fully dissociate into ions in solution. Hydrofluoric acid and ammonia are weak electrolytes, partially dissociating into ions. Ethanol and glucose are nonelectrolytes, dissolving without forming ions.
Explanation:
When classifying solutes as strong electrolytes, weak electrolytes, or nonelectrolytes, we examine how they dissociate in water. A strong electrolyte dissociates completely into ions, a weak electrolyte only partially dissociates, and a nonelectrolyte does not dissociate at all in solution.
Nitric acid (HNO3) is a strong electrolyte because it dissociates completely into H+ and NO3- ions in aqueous solution.Lithium hydroxide (LiOH) is also a strong electrolyte, as it dissociates into Li+ and OH- ions when dissolved in water.Hydrofluoric acid (HF) is a weak electrolyte because it only partially dissociates into H+ and F- ions.Ammonia (NH3), when dissolved in water, forms ammonium (NH4+) and hydroxide (OH-) ions, but because the reaction is reversible and the degree of dissociation is not complete, it is also considered a weak electrolyte.Potassium chloride (KCl) is a strong electrolyte that fully dissociates into K+ and Cl- ions.Ethanol (C2H5OH) and glucose (C6H12O6) are both nonelectrolytes; they dissolve as molecules without forming ions in aqueous solution.If you were asked to add 50 mL of a stock solution to a reaction vessel, which piece of glassware would you use to obtain this volume of solution?
To measure and add 50 mL of a stock solution, a volumetric pipette or graduated cylinder should be used for accuracy and precision in chemical laboratory work, as volumetric flasks are not suitable for measuring out volumes other than their calibration capacity.
Explanation:If you need to add 50 mL of a stock solution to a reaction vessel, the most appropriate and precise piece of glassware to use would be ab this case. When dealing with stock solutions, it's important to use glassware that ensures precision to ensure the final concentrations of your solutions are accurate.
When mixing solutions or reagents, accuracy and precision are crucial, especially in a chemical laboratory or medical environment. For instance, preparing an IV solution with a specific concentration involves accurate measurements of the stock solution and addition of the solvent up to a desired volume. Misjudgements in the volume of stock solution added could alter the final concentration of the solution, which could have significant implications, for example, in medical applications.
What concentration of benzoic acid will have a ph of 4.0? (for benzoic acid, pka=4.20)?
Write a net ionic equation to show how codeine, c18h21o3n, behaves as a base in water.
Actually, the ionic equation for this is a reversible equation since codeine is a weak base. Any weak base or weak acids do not completely dissociate which makes them a reversible process. The ionic equation for this case is:
C18H21O3N + H3O+ <=> C18H21O3NH+ + H2O
The complete combustion of ethanol, c2h5oh(l), to form h2o(g) and co2(g) at constant pressure releases 1235 kj of heat per mole of c2h5oh. write a balanced equation for this reaction.
Since it was specifically stated that heat is released, therefore heat is located on the right side of the reaction equation. Also stating the states of each compound, the complete balanced reaction would be:
C2H5OH (l) + 3 02 (g) -----> 2 CO2 (g) + 3 H2O (g) + 1235 kJ
The balanced combustion reaction will be
[tex]\rm \bold {C_2H_5OH_ (_l_) + 3 0_2 _(_g_) \rightarrow 2 CO_2 _(_g_) + 3 H_2O (g) + 1235 kJ}[/tex]
To write a balanced equation
We will carefully put reactant in the left and product into the right. The energy released will be mentioned in right.The states of reactant and product will also mentioned.Hence, balanced reaction will be
[tex]\rm \bold {C_2H_5OH_ (_l_) + 3 0_2 _(_g_) \rightarrow 2 CO_2 _(_g_) + 3 H_2O (g) + 1235 kJ}[/tex]
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What is the volume of a solution that has a specific gravity of 1.2 and a mass of 185 g?
Is helium a homogeneous or heterogeneous mixture or compound or element?
An element has three naturally occurring isotopes. Use the information below to calculate the weighted average atomic mass of the element, showing both the setup and the final answer for the calculation.
Isotope
Atomic Mass
Percent Abundance
I
39.96 u
99.59%
II
37.96 u
0.065%
III
35.97 u
0.336%
Propane has the formula c3h8. if a sample of propane contains 0.20 moles of c, how many moles of h are present?
Solving this problem only requires the knowledge of stoichiometry. We are given the compound propane C3H8. From the chemical formula, we can see that for every 3 moles of C, there are 8 moles of H. Therefore:
moles of H = 0.20 moles C * (8 moles H / 3 moles C)
moles of H = 0.53 moles H
Answer: In the given sample, 0.533 moles of hydrogen will be present.
Explanation:
We are given a chemical compound having formula [tex]C_3H_8[/tex]
We are provided with 0.20 moles of carbon atoms.
As, for every 3 moles of carbon atoms, 8 moles of hydrogen atoms are present.
So, for 0.20 moles of carbon atoms, [tex]\frac{8}{3}\times 0.20=0.533moles[/tex] of hydrogen atoms will be present.
Hence, in the given sample, 0.533 moles of hydrogen will be present.
The half-life of radioactive element krypton-91 is 10 seconds. if 16 grams of krypton-91 are initially present, how many grams are present after 10 seconds? after 50 seconds?
After one half-life (10 seconds), 8 grams of krypton-91 will remain. After 50 seconds (or 5 half-lives), only 0.5 grams will be left.
Explanation:The question pertains to the concept of half-life in radioactive decay. Half-life is the time taken for half of the radioactive element to decay. With a half-life of 10 seconds, krypton-91 follows an exponential decay pattern. For example, if you start with 16 grams of krypton-91, after one half-life (10 seconds), half of it will decay, leaving you with 8 grams of krypton-91.
After 50 seconds or five half-lives, the remaining krypton-91 would be 16g * (1/2)^5 = 0.5 grams. It is important to note that the radioactive atoms don't disappear but are replaced by their decay products or daughter elements.
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Large molecules that contain carbon and are held together by covalent bonds are called
AgNO3(aq)+NaCl(aq)------>AgCl(s)+NaNO3(aq)
Name the chemical that are dissolved in water
In the reaction AgNO3(aq)+NaCl(aq)------>AgCl(s)+NaNO3(aq), the chemicals dissolved in water are AgNO3, NaCl, and NaNO3.
Explanation:In the chemical reaction AgNO3(aq)+NaCl(aq)------>AgCl(s)+NaNO3(aq), the chemicals that are dissolved in water, also known as aquatic solutions, are AgNO3 (Silver Nitrate) and NaCl (Sodium Chloride) on the reactant side, and NaNO3 (Sodium Nitrate) on the product side. These are shown with the (aq) notation which indicates they are aquatic, or dissolved in water.
The product AgCl (Silver Chloride), shown with the (s) notation, demonstrates that it precipitates, or forms a solid in the solution instead of dissolving in the water.
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Volume measurements will be made using a 25-ml graduated cylinder, which has major scale divisions marked every 2 ml and minor scale divisions every 0.2 ml. what is the estimated uncertainty in the volume measurements
Actually, the uncertainty point is when we have to estimate one place smaller than what the graduated cylinder can accurately read. So in this case, the smallest that we can accurately read is given by the minor marks of 0.2 mL. So if it is in between two minor marks we can only estimate the number, therefore the uncertainty is +/- 0.1ml.
Answer:
The estimated uncertainty is ±0.1 ml
Explanation:
Measurements are usually associated with uncertainty which depends on the instrument used or the procedure employed.
It is given that the graduated cylinder of volume 25 ml has a major scale marked every 2 ml and minor scale with 0.2 ml. In general measurements made with graduated cylinders can be estimated one decimal place beyond the smallest division. In this case, the volume of the liquid can be read to the nearest 0.1 division. Hence, the uncertainty would be ± 0.1 ml.
express 601 kg as an integer
601kg as integers can be written as 601000
Explanation:In mathematics an integer is a number that can be written without a fractional or decimal component. For example: 21, 4, and −2048 are integers; 9.75, 5½, and √2 are not integers. If zero comes in the middle of the integer it is taken into account but if it comes in the start then it is discarded. Similarly we have scientific notations and we know that kg is equal to 1000 so we can write 601 kg as 601000 which fulfills both the definition of integer and scientific notation.
601 kg expressed as an integer remains 601 kg, since it is already a whole number and does not require any conversion.
Explanation:To express 601 kg as an integer, we do not need to apply any conversions, as 601 kg is already an integer value. The term 'integer' refers to a whole number that is not a fraction or a decimal. Therefore, 601 kg is an integer as it stands. It's important to note that when measuring mass or weight, it's beneficial to use units that keep the number manageable and avoid unnecessary complexity, as described with the examples of different weight measurements and their ease based on the unit used.
Calculate the specific heat of a substance when 63j of energy are transferred as heat to an 8.0 g sample to raise it temperature from 314 k to 340 k
The formula for energy or enthalpy is:
E = m Cp (T2 – T1)
where E is energy = 63 J, m is mass = 8 g, Cp is the specific heat, T is temperature
63 J = 8 g * Cp * (340 K – 314 K)
Cp = 0.3 J / g K
Answer : The specific heat of substance is 0.30 J/g.K
Explanation :
Formula used :
[tex]Q=m\times c\times \Delta T[/tex]
or,
[tex]Q=m\times c\times (T_2-T_1)[/tex]
where,
Q = heat = 63 J
m = mass of substance = 8.0 g
[tex]C_w[/tex] = specific heat of substance = ?
[tex]T_1[/tex] = initial temperature = 314 K
[tex]T_2[/tex] = final temperature = 340 K
Now put all the given value in the above formula, we get:
[tex]63J=8.0g\times c\times (340-314)K[/tex]
[tex]c=0.30J/g.K[/tex]
Therefore, the specific heat of substance is 0.30 J/g.K