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2 years ago

What is the density of 2.5 g of gaseous sulfur held at 130. kPa and 10.0 degrees Celsius?

2 answers:

7 0

Assuming that the gaseous sulfur will behave as an ideal gas,
(PV)/T = constant
(P₁V₁)/T₁ = (P₂V₂)/T₂
(101.3 x 22.4)/(273) = (130 x V₂)/(283)
V₂ = 18.1 L
density = mass/volume
density = 2.5/18.1
= 0.14 g/L

worty [1.4K]2 years ago

4 0

Use ideal gas equation: pV = nRT

Now pass n to mass: n = mass / MM .... [MM is the molar mass]

pV = [mass/MM]*RT =>mass/V = [p*MM] / RT and mass / V = density

p= 130 kPa = 130,000 Pa = 130,00 joule / m^3
T = 10.0 ° + 273.15 = 283.15 k
MM of sulfur (S) = 32 g/mol = 32000 kg/mol

density = 130,000 Pa * 32000kg/mol / [8.31 joule / mol*k * 283.15 k] = 1.77*10^6 kg/m^3 = 1.77 g/L ≈ 1.8 g/L

Then, I do not get any of the option choices.

Is it possbile that the pressure is 13.0 kPa instead 130. kPa? If so the answer would be 18 g/L

Note that the mass is not used. You do not need it unless you are asked for the volume, which is not the case.

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The following reaction shows sodium hydroxide reacting with sulfuric acid.

amm1812

its answer is 35.5 grams

6 0

2 years ago

Read 2 more answers

Be sure to answer all parts. ΔH o f of hydrogen chloride [HCl(g)] is −92.3 kJ/mol. Given the following data, determine the ident

Akimi4 [234]

Answer:

NH₄Cl(s) → NH₃(g) + HCl(g)

ΔH°rxn 74,89 kJ/mol

Explanation:

The change in enthalpy of formation (ΔHf) is defined as the change in enthalpy in the formation of a substance from its constituent elements. For HCl(g):

(1) ¹/₂H₂(g) + ¹/₂ Cl₂ → HCl(g) ΔH = -92,3 kJ/mol

It is possible to sum ΔH of different reactions to obtain ΔH of a global reaction (Hess's law).

For the reactions:

(2) N₂(g) + 4H₂(g) + Cl₂(g) → 2NH₄Cl(s) ΔH°rxn = −630.78 kJ/mol

(3) N₂(g) + 3H₂(g) → 2NH₃(g) ΔH°rxn = −296.4 kJ/mol

The sum of -(2) + (3) gives:

-(2) 2NH₄Cl(s) → N₂(g) + 4H₂(g) + Cl₂(g) ΔH°rxn = +630.78 kJ/mol

(3) N₂(g) + 3H₂(g) → 2NH₃(g) ΔH°rxn = −296.4 kJ/mol

-(2) + (3) 2NH₄Cl(s) → 2NH₃(g) + H₂(g) + Cl₂(g)

ΔH°rxn = +630.78 kJ/mol −296.4 kJ/mol = +334,38 kJ/mol

Now, the sum of -(2) + (3) + 2×(1)

-(2) + (3) 2NH₄Cl(s) → 2NH₃(g) + H₂(g) + Cl₂(g) ΔH°rxn = +334,38 kJ/mol

2×(1) H₂(g) + Cl₂(g)→ 2HCl(g) ΔH = 2×-92,3 kJ/mol

-(2) + (3) + 2×(1) 2NH₄Cl(s) → 2NH₃(g) + 2HCl(g)

ΔH°rxn = +334,38 kJ/mol + 2×-92,3 kJ/mol = 149,78 kJ/mol

The reaction of:

NH₄Cl(s) → NH₃(g) + HCl(g)

Has ΔH°rxn = 149,78kJ/mol / 2 = 74,89 kJ/mol

I hope it helps!

8 0

1 year ago

ANSWER Soon!!

vichka [17]

Answer:

1. 0.1683 mol

2. 1.191 g

3. 0.02695 mol

4. Na₂Cl₃

5. The empirical formula obtained is not correct. This is likely due to experimental errors, since much precision was required (the masses are determined in thousandths of grams).

Explanation:

1. How many moles of elemental sodium were used in the reaction?

Since all of the solid sodium is used up by the reaction, you can cancluate the number of moles of elemental sodium used dividing the mass by the molar mass:

number of moles = mass in grams / atomic mass

mass in grams = 0.3870 g (given)

atomic mass = 22.990 g/mol

number of moles = 0.3870 g / 22.990 g/mol = 0.1683 mol

2. What is the mass of chlorine gas used in the reaction?

a) Mass of chlorine gas introduced in the flask = mass of the stoppered flask after filling it with chlorine gas - mass of the empty flask with the sopper

Mass of chlorine gas introduced = 158.1743g - 156.1870g = 1.9873 g

b) Mass of chlorine gas unreacted = 0.7962 g (given)

c) Mass of chlorine gas used = mass of chorine gas introduced in the flask - mass of chlorine gass un reacted

Mass of chlorine gas used = 1.9873g - 0.7962g = 1.1911g

3. How many moles of chlorine were used in the reaction?

molar mass of chlorine gas, Cl₂ = 2 × 35.453 g/mol = 70.906 g/mol

number of moles = mass in grams / molar mass = 1.911g / 70.906g/mol = 0.02695 mol

4. What is the empirical formula of sodium chloride based on the experimental data?

Divide the number of moles of each element by the smalles number of moles:

Na: 0.01683 / 0.01683 = 1

Cl = 0.02695 / 0.01683 = 1.6

Multiply by 2 to obtain whole numbers:

Na = 2

Cl = 3.2 = 3

Empirical formula Na₂Cl₃

5. Was the empirical formula you obtained correct using the chemists data correct? Why?

No, the empirical formula you obtained using the chemists data is not correct, because the correct empirical formula of sodium chloride is NaCl.

That is, there is 1 atom of sodium per every atom of chlorine in one chemical formula of NaCl, but that is not reflected by the empirical formula Na₂Cl₃.

That is a demostration of big experimental errors. You can speculate that the errors are likely due to problems of procedure collecting the gas or errors in measuring the masses.

As you see, the masses are measured to thousandths of grams, which requires much precision; thus smalls absolute errors could produce huge relative errors.

7 0

2 years ago

ANSWER FOR: How did the lab activities help you answer the lesson question "How do the processes of conduction, convection, and

Alenkinab [10]

Answer: Conduction, convection, and radiation move energy from the Sun to Earth and throughout Earth.

Without more information about the experiment itself, I would choose the above answer as correct. All the other statements are correct, however none of them relates to the earth distribution processes on Earth. The last statement does.

8 0

1 year ago

Read 2 more answers

Are carboxylic acids of more than 10 carbons more soluble in polar or nonpolar solvents? Explain.

lina2011 [118]

Answer: Non polar solvents

Explanation:

Since with increasing the size of alkyl group hydrophobic nature increases and solubility in polar solvents decreases .

Hence Carboxylic acids with more than 10 carbon atoms, solubility is more in non polar solvents.

8 0

1 year ago