How many grams of h3po4 are in 521 ml of a 9.30 m solution of h3po4?

Answers

Answer 1

Final answer:

To find the mass of H3PO4 in 521 ml of a 9.30 M solution, convert the volume to liters, multiply by the molarity to get moles, and then convert moles to grams using the molar mass of H3PO4. Approximately 474.6434 grams of H3PO4 are in 521 ml of a 9.30 M solution.

Explanation:

The student has asked how many grams of H3PO4 are in 521 ml of a 9.30 M solution. To find this, we use the formula:

Calculate the number of moles of H3PO4 by multiplying the molarity (M) of the solution by the volume of the solution in liters.

Convert the moles of H3PO4 to grams using the molar mass of H3PO4 (which is approximately 98 g/mol).

First, we convert 521 ml to liters (521 ml * 0.001 L/ml = 0.521 L). Next, we calculate the moles of H3PO4 (9.30 M * 0.521 L = 4.8433 mol). Finally, we multiply the number of moles by the molar mass of H3PO4 to get the mass in grams (4.8433 mol * 98 g/mol = 474.6434 g).

Therefore, there are approximately 474.6434 grams of H3PO4 in 521 ml of a 9.30 M solution.


Related Questions

Pressure and volume are inversely related. When the pressure on a gas is doubled, what happens to the volume

Answers

When the pressure is doubled the volume will be half as the pressure increase due to the frequent collision of molecules so with the decrease in volume the pressure will increase as Boyles law

Answer: Volume decreases to half of original volume

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

[tex]P\propto \frac{1}{V}[/tex]     (At constant temperature and number of moles)

[tex]{P_1V_1}={P_2V_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = p

[tex]P_2[/tex] = final pressure of gas = 2p

[tex]V_1[/tex] = initial volume of gas = v

[tex]V_2[/tex] = final volume of gas = ?

Now put all the given values in the above equation, we get the final pressure of gas.

[tex]{p\times v}=2p\times V_2[/tex]

[tex]V_2=\frac{v}{2}[/tex]

Therefore, the final volume of the gas will become half of initial volume.

A gas cylinder contains exactly 15 moles of oxygen gas (O2). How many molecules of oxygen are in the cylinder? 4.01 × 1022 molecules 6.02 × 1023 molecules 9.03 × 1024 molecules 2.89 × 1026 molecules

Answers

Moles are used conveniently in chemistry especially in stoichiometric calculations involving reactions. The unit of mole is a collective term that holds 6.022×10^23 particles. These particles is a general term for any small units of matter including molecules, atoms and sub-particles. This ratio of 6.022×10^23 particles to 1 mole is known to be the Avogadro's number. Its exact number is actually 6.0221409×10^23. We use this constant in our stoichiometric calculation as follows:

15 moles oxygen * (6.022×10^23 molecules/ 1 mole oxygen) = 9.033×10^24 molecules of oxygen

Answer:

answer in picture    It's B

Explanation:

Compare bond lengths in butane and t butylcyclohexane

Answers

The bond angles of butane and tert butylcyclohexane are different. First, butane is an alkane that has four carbon atoms and its molecular geometry is that of a straight line. While tert butylcyclohexane is a cycloalkane that forms a ring like structure with one carbon atom attached with a tert butyl branch. Second, the tert butylcyclohexane is shorter in bond length and is strained due to the branched alkyl attached to the cycloalkyl compared to the butane that has no steric hindrance.

Answer:

The lengths of the C-C bonds increases with the decrease in the resistance of said bond, for example, a triple bond has a shorter length than in the case of a single bond. Butane has the single bonds and the CC bond is hybridized with sp3 hybridization, however in the butylcyclohexane structure the CC bond is also sp3 and the angle is 120°, however the angle shown is equal to 109.5°, so there is a certain angular tension and it is very unstable with respect to butane

Explanation:

When a colorless aqueous solution of lead nitrate is combined with a colorless aqueous solution of sodium iodide a bright yellow precipitate is formed. what is the chemical formula for the precipitate?

Answers

Pb(NO₃)₂(aq) + 2NaI(aq) → 2NaNO₃(aq) + PbI₂(s)

PbI₂ (the precipitate)

The chemical formula of the bright yellow precipitate is PbI₂ (lead iodide).

What is the balanced chemical equation?

A chemical equation is the representation of a chemical reaction which consists of reactants participating, formed products, and an arrow indicating the direction of the chemical reaction.

The equation that has the number of atoms of substances equal on either side of the chemical equation is known as a balanced chemical equation.

The law of conservation of mass has to be followed by a balanced chemical equation, according to which, the total mass of the elements on the reactant side must be equal to the total mass of elements on the product side.

The chemical equation of the reaction of lead nitrate and an aqueous solution of sodium iodide:

[tex]Pb(NO_3)_2(aq) + 2NaI \longrightarrow 2NaNO_3(aq) + PbI_2 (s)[/tex]

The bright yellow precipitate formed in the above chemical reaction has the chemical formula PbI₂.

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Phosphorus has three unpaired electrons and hydrogen has one unpaired electron this means that_____ equivalents of hydrogen can react with ______ equivalents of phosphorus.

Answers

Three equivalents of hydrogen
One equivalent of phosphorus

Three equivalents of hydrogen can react with one equivalent of phosphorus to form compounds like phosphine, where each hydrogen atom forms a bond with one of the unpaired electrons of phosphorus.

Phosphorus typically has three unpaired electrons and hydrogen has one unpaired electron, which means that three equivalents of hydrogen can react with one equivalent of phosphorus. For instance, in the formation of phosphine, PH₃, three hydrogen atoms will combine with one phosphorus atom, each hydrogen providing one electron to form a single bond with phosphorus. Since phosphorus has three unpaired electrons available, it is able to form three single bonds with three hydrogen atoms, resulting in the phosphine compound.

What is the mass loss of the nucleus, in u, upon emission of this gamma ray? -g?

Answers

Suppose that the energy of the emitted gamma ray photon is given the symbol "e" in units of Kev
Now, we use Einstein's equation (E = mc² ) where c is the speed of light (3 x 10^8 m/sec)

substituting in this equation, we find that:
e = mass x c² 
mass = e / c² 
Now to get the corresponding value in atomic mass unit:
1 eV/c² = 1/(931.46 * 10^6) amu 
Therefore,
The mass in amu = e x  1/(931.46 * 10^6) atomic mass unit


 

Which of the following does not apply to obtaining a career in chemistry?

•There is only one path to get a career in chemistry.
•There are many paths to getting a career in chemistry.
•Start by talking to teachers and counselors about different career options.
•Talk to students already enrolled in the programs that you are interested in.

Answers

a, the first one, does not apply

ANSWER: There is only one path to get a career in chemistry.

EXPLANATION: There are many paths of career in Chemistry and not only one. It would not be justified to say that there is only one path to get a career in chemistry. Careers in chemistry would include Biochemistry, Forensic Scientist, Research Scientist, Chemical Engineer, Chemical Plant Operator and even a Science teacher or Chemistry professor in a school or university.

Calculate the vapor pressure of a solution containing 27.2 g of glycerin (c3h8o3) in 132 ml of water at 30.0 ?c. the vapor pressure of pure water at this temperature is 31.8 torr. assume that glycerin is not volatile and dissolves molecularly (i.e., it is not ionic) and use a density of 1.00 g/ml for the water.

Answers

This problem is to apply Roult's Law.

Roult's Law states that the vapor pressure, p, of a solution of a non-volatile solute is equal to the vapor pressure of the pure solvent, Po solv, times the mole fraction of the solvent, Xsolv

p = Xsolv * Po sol

X solv = number of moles of solvent / number of moles of solution

The solvent is water and the solute (not volatile) is glycerin.

Number of moles = mass in grams / molar mass

mass of water = 132 ml * 1 g/ml = 132 g

molar mass of water = 18 g/mol

=> number of moles of water = 132 g / 18 g/mol = 7.33333 mol

mass of glycerin = 27.2 g

molar mass of glycerin:, C3H8O3: 3 * 12 g/mol + 8 * 1 g/mol + 3*16 g/mol = 92 g/mol

number of moles of glycerin = 27.2g / 92 g/mol = 0.29565

total number of moles = 7.33333 moles + 0.29565 moles = 7.62898 moles

=> X solv = 7.33333 / 7.62898 = 0.96125

=> p = 0.96125 * 31.8 torr ≈ 30.57 torr ≈ 30.6 torr.

Answer: 30.6 torr

Final answer:

To find the vapor pressure of the glycerin solution, calculate the moles of glycerin and water, determine the mole fraction of water, and apply Raoult's law using the vapor pressure of pure water at the specified temperature.

Explanation:

To calculate the vapor pressure of the solution containing glycerin in water, we will use Raoult's law, which states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent. The first step is to calculate the number of moles of glycerin (C3H8O3) by using its molar mass (92.09 g/mol), and then calculate the number of moles of water using its given density (1.00 g/mL) to convert the volume to mass and then to moles with its molar mass (18.015 g/mol).

Once we have both amounts in moles, we can calculate the mole fraction of water and apply Raoult's law to find the new vapor pressure of the solution, knowing the vapor pressure of pure water at the given temperature (30.0 °C) is 30.6 Torr.

The generic metal a forms an insoluble salt ab(s) and a complex ac5(aq). the equilibrium concentrations in a solution of ac5 were found to be [a] = 0.100 m, [c] = 0.0110 m, and [ac5] = 0.100 m. determine the formation constant, kf, of ac5.

Answers

Assuming that the reaction from A and C to AC5 is only one-step (or an elementary reaction) with a balanced chemical reaction of:

A + 5 C  --->  AC5

Therefore the formation constant can be easily calculated using the following formula for formation constant:

Kf = product of products concentrations / product of reactants concentration

Kf = [AC5] / [A] [C]^5                    

---> Any coefficient from the balanced chemical reaction becomes a power in the formula

Substituting the given values into the equation:

Kf = 0.100 M / (0.100 M) (0.0110 M)^5

Kf = 6,209,213,231

or in simpler terms

Kf = 6.21 * 10^9                  (ANSWER)

Why is butane in the lighter a liquid yet the butane in the buret is a gas?

Answers

This because when it is in the lighter its is not in contact with the outside environment so the  pressure does not act on it or act in very small amount while in buret it is open from the head and thus pressure acts on it and which cause its vapor pressure to increase more than atmospharic pressure and cause it to change state 
This is because if the vapor pressure rise above atmospharic pressure then it reaches to its boiling point and change state to gas thats why in buret due to pressure difference its is in gas state

The main difference between heat and temperature is the temperature is Solely dependent on the

Answers

Velocity of molecules in the body

Answer:the answer is the Velocity of molecules in the body

Explanation:hope this helped have a great day

What are three physical properties often analyzed to identify covalent compounds?

Answers

Covalent compounds are compounds in which nonmetal atoms are bond and share electrons. 
In order to identify if a compound is covalent it must satisfy the following physical properties:
1. Covalent compounds generally have much lower melting and boiling points than ionic compounds.
2. Covalent compounds have soft or brittle solid forms.
3. Covalent compounds have poor electrical and thermal conductivity.
Final answer:

Covalent compounds can be identified by their low melting and boiling points, non-conductivity of electricity, and solubility in nonpolar solvents.

Explanation:

Covalent compounds, also referred to as molecular compounds, are usually identified through several physical properties such as their low melting and boiling points, non-conductivity of electricity, and their solubility in nonpolar solvents.

Low melting and boiling points: Covalent compounds don't have strong intermolecular forces, resulting in low melting and boiling points.

Non-conductivity of electricity: Since most covalent compounds don't dissociate into ions in solutions, they typically do not conduct electricity.

Solubility in nonpolar solvents: Due to the 'like soluble like' rule, covalent compounds tend to be soluble in nonpolar solvents rather than polar solvents such as water.

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How much energy would be released if 1.0 g of material were completely converted into energy?

Answers

Using this formula E = mc2, which is the formula formulated by Albert Einstein to get the energy where E is the units of energy, m is the mass and c is the speed of light. We can say that 1 g is equivalent to 0.001 kg. The speed of light is 38. Substituting these values to the formula, we can get 90 terajoules.

Final answer:

The energy released from converting 1.0 g of mass into energy is 9 × 10¹³ joules (J), using the equation E = mc², where c is the speed of light.

Explanation:

According to Einstein's famous equation E = mc², where E represents energy, m is mass, and c is the speed of light in a vacuum, the energy released from completely converting 1.0 g of mass into energy is tremendously large. Since the speed of light, c, is approximately 3 × 10⁸ meters per second, and the mass m is 1.0 g (which is 1/1000 of a kilogram), the calculation is E = (1.0 g / 1000) × (3 × 10⁸ m/s)². This results in an energy release of 9 × 10¹³ joules (J), which is equivalent to about twice the energy released by the atomic bomb dropped on Hiroshima.

Calculate the equilibrium constant k for the isomerization of glucose-1-phosphate to fructose-6-phosphate at 298 k. express your answer numerically using two significant figures. hints

Answers

We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.

glucose-1-phosphate⟶glucose-6-phosphate          ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate          ΔG∘=−1.67 kJ/mol

Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.

glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate 

In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.

ΔG°,total = −7.28 kJ/mol  + 1.67 kJ/mol = -5.61 kJ/mol

Then, the equation to relate ΔG° to the equilibrium constant K is

ΔG° = -RTlnK, where R is the gas constant equal to 0.008317 kJ/mol-K.
-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)
lnK = 2.2635
K = e^2.2635
K = 9.62


The equilibrium constant for isomerization reaction is [tex]\boxed{9.615}[/tex]

Further Explanation:

The standard Gibbs free energy change in a reaction [tex]\left( {{{\Delta {\text{G}}_{{\text{rxn}}}^{{^\circ }}} \right)[/tex] is the difference of sum of the standard free energies of formation of product molecules and sum of standard free energies of formation of reactant molecules at the standard conditions. The formula used to calculate the value of standard Gibbs free energy  change for a reaction [tex]\left( {{{\Delta {\text{G}}_{{\text{rxn}}}^{{^\circ }}} \right)[/tex] is as follows:

[tex]\Delta\text{G}_{\text{rxn}}^{\circ}=\sum\text{n}\Delta\text{G}_{\text{f}(\text{products})}^{\circ}-\sum\text{m}\Delta\text{G}_{\text{f}(\text{reactants})}^{\circ}[/tex]

Here, n is the stoichiometric coefficients of products, and m are the stoichiometric coefficients of reactants in a balanced chemical equation.

The formula to determine the relationship between change in standard Gibbs free energy [tex]\left( \Delta{\text{G}^{\circ}} \right)[/tex] and equilibrium constant [tex]\left({\text{K}}\right)[/tex] is given as follows:

[tex]{\Delta }}{{\text{G}}^{{^\circ }}} = - {\text{RTlnK}}[/tex]       ......(1)

Here,

[tex]\Delta{\text{G}^{\circ}[/tex] is the standard Gibbs free energy change.

[tex]{\text{R}[/tex] is the gas constant.

[tex]{\text{T}}[/tex] is the temperature in Kelvin.

[tex]{\text{K}}[/tex] is the equilibrium constant.

The isomerization of glucose-1-phosphate to fructose-6-phosphate occurs in 2 steps:  

The reaction of step 1 is as follows:

[tex]{\text{glucose - 1 - phosphate}} \to {\text{glucose - 6 - phosphate}}[/tex]

                                       ......(2)

[tex]\Delta{\text{G}^{\circ}_{1}[/tex] for equation (2) is  [tex]- 7.28\;{\text{kJ/mol}}[/tex]

The reaction of step 2 is as follows:

[tex]{\text{fructose - 6 - phosphate}} \to {\text{glucose - 6 - phosphate}}[/tex]

                                                   ......(3)

[tex]\Delta{\text{G}^{\circ}_{2}[/tex] for equation (3) is  [tex]- 1.67\;{\text{kJ/mol}}[/tex]

Reverse the reaction of step 2.

[tex]{\text{glucose - 6 - phosphate}} \to {\text{frutcose - 6 - phosphate}}[/tex]

                                                ......(4)

[tex]\Delta{\text{G}^{\circ}_{3}[/tex] for equation (4) is [tex]+ 1.67\;{\text{kJ/mol}}[/tex]

Add equation (1) and (3) to get the final equation.

[tex]{\text{glucose - 1 - phosphate}} \to {\text{frutcose - 6 - phosphate}}[/tex]

To calculate [tex]\Delta {\text{G}}_{{\text{rxn}}}^{^\circ }}[/tex], add [tex]\Delta{\text{G}^{\circ}_{1}[/tex] and [tex]\Delta{\text{G}^{\circ}_{3}[/tex] as follows:  

[tex]\Delta{\text{G}^{\circ}_{\text{rxn}}=\Delta{\text{G}^{\circ}_{1}+\Delta{\text{G}^{\circ}_{3}[/tex]                       ......(5)

Substitute [tex]- 7.28\;{\text{kJ/mol}}[/tex] for [tex]\Delta{\text{G}^{\circ}_{1}[/tex] and [tex]+ 1.67\;{\text{kJ/mol}}[/tex] for [tex]\Delta{\text{G}^{\circ}_{3}[/tex] in equation (5).

[tex]\begin{aligned}\Delta {\text{G}}_{{\text{rxn}}}^{{^\circ }} &=  - 7.28\;{\text{kJ/mol + }} + 1.67\;{\text{kJ/mol}}\\{\text{}}&= - 5.61\;{\text{kJ/mol}}\\\end{aligned}[/tex]

For equilibrium constant (K), rearrange equation (1)

[tex]{\text{K}}={\text{e}}\frac{-\Delta{\text{G}}^{\circ}}{\text{RT}}[/tex]    ......(6)

Substitute [tex]- 5.61\;{\text{kJ/mol}}[/tex] for [tex]\Delta{\text{G}^{\circ}[/tex],[tex]8.314\;{\text{J/mol}} \cdot {\text{K}}[/tex] for R and [tex]298\;{\text{K}}[/tex] for T in equation (6)

[tex]\begin{aligned} {\text{K}}&= {{\text{e}}^{\frac{{ - \left( { - 5.61\;{\text{kJ/mol}}} \right)}}{{\left( {8.314\;{\text{J/mol}} \cdot {\text{K}}} \right)\left( {\frac{{{\text{1J}}}}{{1000{\text{kJ}}}}} \right)\left( {298\;{\text{K}}} \right)}}}}\\&= {{\text{e}}^{2.2634}}\\&= 9.615\\\end{aligned}[/tex]

The equilibrium constant for the reaction is 9.615.

Learn more:

1. The change in standard gibbs free is for a reaction: https://brainly.com/question/10838453

2. Determination of the equilibrium constant for pure water: https://brainly.com/question/3467841

Answer details:

Grade: Senior Secondary School

Subject: Chemistry

Chapter: Chemical Equilibrium

Keywords: Standard Gibbs free energy, equilibrium, constant, glucose-1-phosphate and fructose-6-phosphate.

How Does The Modern Periodic Table Arrange Elements? By Atomic Mass By Atomic Number
By Number Isotopes
By Electron Mass

Answers

By increasing Atomic number

The molar mass of carbon dioxide is 44.0 g/mol. a mass of 150.0 grams of carbon dioxide is equivalent to how many moles?

A) 3.00 mol

B) 3.41 mol

C) 29.3 mol

D) 106 mol

Answers

M(CO₂)=44.0 g/mol
m(CO₂)=150.0 g

n(CO₂)=m(CO₂)/M(CO₂)

n(CO₂)=150.0/44.0=3.41 mol

B) 3.41 mol

The molar mass of carbon dioxide is 44.0 g/mol. a mass of 150.0 grams of carbon dioxide is equivalent to 3.41 mol of carbon dioxide.

What is mole?

In the International System of Units, the mole is the unit of substance quantity (SI). How so many elementary units of a certain substance are present in an item or sample is determined by the quantity of that material. There are precisely 6.022×10²³ elementary entities in a mole.

For instance, although having differing volumes and weights, 10 moles containing water along with ten moles of mercury both contain the same quantity of material, and the mercury includes exactly 1 atom for every molecule of water.

M(CO₂)=44.0 g/mol

m(CO₂)=150.0 g

n(CO₂)=m(CO₂)/M(CO₂)

n(CO₂)=150.0/44.0=3.41 mol

Therefore, 3.41 mol of carbon dioxide is there.

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Determine the number of 3s electrons in na.

Answers

Use the atomic number (Z) of the element and determine the electron configuration.

Z for Na is 11 => 11 electrons.

=> electron configuration = 1s2 2s2 2p6 3s

So, there you can count 5 s electrons: 2 in 1s, 2 in 2s and 1 in 3s).

The questions asks only for the 3s electrons, so the answer is one 3s electron.

Calculate the mass of water produced when 1.92 g of butane reacts with excess oxygen.

Answers

c4h10+6.5o2=4co2+5h2o
moles of butane=1.92/58=0.0331 moles
moles of water=0.1655 moles\
as the butane and water has 1 is to 5 molar ratio
0.1655=mass/18
mass=2.98 g
mass of water produced = 2.98 g

The addition of 435.2 j of heat is required to raise the temperature of 3.4 g of olive oil from 21?c to 85?c. what is the specific heat of the olive oil?

Answers

use this formula Q=mcΔT
                          435.2   =(3.4)(c)(85-21)
                          435.2   =217.6c

                                  c  =2 J/kg.c°

The specific heat of olive oil is 2 J/g °C'

From the question,

We are to determine the specific heat of olive oil

From the formula

Q = mcΔT

Where Q is the quantity of heat

m is the mass of substance

c is the specific heat of substance

ΔT is the change in temperature

From the given information

Q = 435.2 J

m = 3.4 g

ΔT = 85 °C - 21 °C = 64 °C

Putting the above parameters into the formula, we get

435.2 = 3.4 × c × 64

435.2 = 217.6c

∴  c = 435.2 ÷ 217.6

c = 2 J/g °C

Hence, the specific heat of olive oil is 2 J/g °C

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55 kg of liquefied natural gas (lng) are stored in a rigid, sealed 0.17 m3 vessel. in this problem, model lng as 100% methane. due to a failure in the cooling/insulation system, the temperature increases to 200 k, which is above the critical temperature; thus, the natural gas will no longer be in the liquid phase.

Answers

The pressure in the vessel after the temperature increase is approximately 33.65 MPa.

We are given a scenario where liquefied natural gas (LNG) stored in a rigid, sealed vessel experiences a temperature increase beyond its critical point, causing it to transition from a liquid to a gas phase. We need to find the final pressure in the vessel using the ideal gas law.

2. Modeling the system:

We treat the LNG as pure methane ([tex]CH_4[/tex]) for simplification.

We assume the system behaves like an ideal gas, meaning it follows the ideal gas law.

3. Setting up the equation:

The ideal gas law relates pressure (P), volume (V), number of moles (n), gas constant (R), and temperature (T) through the equation:

PV = nRT

4. Identifying known and unknown values:

V: 0.17 m³ (volume of the vessel)

R: 8.314 J/(mol·K) (universal gas constant)

T: 200 K (final temperature)

P: Unknown (pressure we need to solve for)

5. Converting mass of LNG to moles:

Molar mass of methane ([tex]CH_4[/tex]): 16.04 g/mol

Mass of LNG (m): 55 kg = 55,000 g

Number of moles (n):

n = m / molar mass

n = 55,000 g / 16.04 g/mol

n ≈ 3433 mol

6. Solving for pressure:

Plug the known values into the ideal gas law and solve for P:

P = (n * R * T) / V

P = (3433 mol * 8.314 J/(mol·K) * 200 K) / 0.17 m³

P ≈ 33,647,247 Pa

7. Converting units and expressing final answer:

Convert pressure from Pascal (Pa) to Megapascal (MPa):

P = 33,647,247 Pa * (1 MPa / 1,000,000 Pa)

P ≈ 33.65 MPa

The question probable may be:

55 kg of liquefied natural gas (lng) are stored in a rigid, sealed 0.17 m3 vessel. in this problem, model lng as 100% methane. due to a failure in the cooling/insulation system, the temperature increases to 200 k, which is above the critical temperature; thus, the natural gas will no longer be in the liquid phase. What would be  pressure in the vessel after the temperature increase

When hydrochloric acid and zinc were combined, the flask was sealed and thus no gas could escape from the flask. why did the pressure decrease as the temperature decreased?

Answers

When hydrochloric acid and zinc were combined, the salt zinc chloride and hydroge gas are being produced in the flask. The flask was said to be sealed and, therefore, no gas could escape from the flask. As the temperature decreases, the pressure inside would decrease since the gas would have lower kinetic energy and would have less force to exert on the walls of the flask. This is also evident from the equation PV = nRT. As we can see pressure and temperature are directly related which means as one decreases, the other decreases as well or as one increases, the other would increase accordingly.

Explanation:

According to the the ideal gas law, PV= nRT.

This means that pressure is directly proportional to temperature.

So, when HCl and zinc are combined in a flask then it will lead to the formation of zinc chloride and hydrogen gas.

The reaction equation will be as follows.

             [tex]HCl + Zn \rightarrow ZnCl_{2} + H_{2}[/tex]

Since, the flask is sealed hence, hydrogen gas will not be able to move out of the flask.

So, when it will behave ideally then due to directly proportional relation between pressure and temperature there will occur a decrease in temperature with decrease in pressure.

Calculate the pH if the pOH is 2.8

Answers

14 - 2.8 = 11.2

because the pH and pOH added together equal 14

If the pOH of a solution is 2.8, you subtract it from 14 to find the pH, resulting in a pH of 11.2.

To calculate the pH from a given pOH, we can use the relationship that the sum of the pH and pOH is equal to 14 at 25 °C (298 K). If the pOH is 2.8, then we can find the pH by subtracting the pOH from 14:

pH = 14 - pOH

pH = 14 - 2.8

pH = 11.2

Therefore, if the pOH of a solution is 2.8, the pH is 11.2.

a 0.175 m weak acid acid solution has ph of 3.25 find ka for the acid

Answers

The acid dissociation constant or Ka is a value used to measure the strength of a specific acid in solution. For a general dissociation of an acid solution,

HA = H+ + A-

we express Ka as follows:

Ka = [H+] [A-] / [HA]

Where the terms represents the concentrations of the acid and the ions. Assuming that the weak acid in the problem is HA, we first calculate for the concentration of H+ from the pH.

pH = - log [H+]
3.25 = - log [H+]
[H+] = 0.0005623 M

By the ICE table, we can calculate the equilibrium concentrations,
        HA      =      H+            +        A-
I      0.175           0                         0
C      -x               +x                      +x
 --------------------------------------------------
E  .174438    0.0005623       0.0005623

Ka = (0.0005623) (0.0005623) / .174438
Ka = 1.81x10^-6
Final answer:

To find the Ka value for the weak acid, we can use the given pH and concentration of the acid solution. The Ka value is calculated using the equation Ka = ([H3O+][A-])/[HA].

Explanation:

To find the value of Ka for the weak acid, we can use the given pH and concentration of the acid solution. We know that the pH is a measure of the concentration of H3O+ ions in a solution, so we can use the pH to calculate the [H3O+] concentration. From the given pH of 3.25, we can determine that the [H3O+] concentration is 10^(-pH). So, [H3O+] = 10^(-3.25) M.

Now, the equilibrium equation for the dissociation of the weak acid is HA(aq) + H2O(l) -> H3O+(aq) + A-(aq). Since we know the [H3O+] concentration, we can assume that the [HA] concentration is equal to the [H3O+] concentration. So, [HA] = [H3O+] = 10^(-3.25) M.

The Ka value is calculated using the equation Ka = ([H3O+][A-])/[HA]. Substituting the given values, Ka = (10^(-3.25)^2)/10^(-3.25). Simplifying this expression gives us the value of Ka for the acid.

________ is the science of obtaining reliable measurements from photographs.

Answers

The answer is photogrammetry.

What gas was produced by the decomposition of hydrogen peroxide? what happened when the smoldering toothpick came into contact with the gas? b boldi italicsu underline bulleted list numbered list superscript subscript?

Answers

The gas that is produced from the decomposition of hydrogen peroxide would be oxygen. Hydrogen peroxide (H2O2) would decompose into water (H2O) and oxygen (O2) where water is in a liquid state and oxygen is in the gas state at STP. When the smoldering toothpick would come in contact with the oxygen that is produced, it would start burning again or a combustion reaction would happen.

Find the missing part of this equation

Answers

Missing part will be 100cm/m, and 6500 respectively

Which of the following is an indication that a substance has undergone a chemical change?

No new product has been formed.

The color of the substance has not changed.

The original constitute has not changed.

The molecular structure has changed.

Answers

The correct option is : THE MOLECULAR STRUCTURE HAS CHANGED.
A chemical change is a type of change in which a new product is formed as a result of reacting with another substance. When a chemical change occur, the molecular structure of the reactant will be altered and an entirely new product will be formed.
All the other options given are examples of physical change.

Answer:

D.

The molecular structure has changed.

Explanation:

I got it right on Plato.

What acid and what base would you choose to prepare the salt potassium perchlorate (kclo4)?

Answers

HClO₄ + KOH → KClO₄ + H₂O

HClO₄ - perchloric acid
KOH - potassium hydroxide

Answer : The perchloric acid and potassium hydroxide base is used to prepare the salt of potassium perchlorate, [tex](KClO_4)[/tex]

Explanation :

when the perchloric acid react with the potassium hydroxide as a base to form a salt of potassium perchlorate, [tex](KClO_4)[/tex]

The balanced chemical reaction will be,

[tex]HClO_4+KOH\rightarrow KClO_4+H_2O[/tex]

By the stoichiometry, we can say that 1 mole of perchloric acid react with the 1 mole of potassium hydroxide base to give 1 mole of potassium perchlorate and 1 mole of water as a product.

Hence, the perchloric acid and potassium hydroxide base is used to prepare the salt of potassium perchlorate, [tex](KClO_4)[/tex]

A sample of hydrated sodium thiosulfate has a mass of 6.584 g. After it is heated, it has a mass of 4.194 g. What is the percentage by mass of water in the hydrate?

Answers

Na₂S₂O₃·xH₂O → Na₂S₂O₃ + xH₂O

w(H₂O)=100m(H₂O)/m₀

m(H₂O)=m₀-m₁

w(H₂O)=100(m₀-m₁)/m₀

w(H₂O)=100(6.584-4.194)/6.584=36.30%




36.30%

[only 20 chars]

a) Diamond and graphite are two different forms of pure elemental carbon with densities of 3.51 g/cc and 2.25 g/cc respectively. What volume would be occupied by a 0.50 g diamond? What volume would be occupied by a 0.50 g piece of graphite?

Answers

Density can be calculated using the following rule:
density = mass / volume
Therefore,
volume = mass / density

For diamond:
we have mass = 0.5 grams and density = 3.51 g / cm^3
Substituting in the rule, we can calculate the volume of diamond as follows:
volume = 0.5 / 3.51 = 0.14245 cm^3

For graphite:
we have mass = 0.5 grams and density = 2.25 g / cm^3
Substituting in the rule, we can calculate the volume of graphite as follows:
volume = 0.5 / 2.25 = 0.2223 cm^3
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