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zhuklara [117]
3 years ago
6

What is the molarity of a solution of Cu2SO4 if 250.0 mL of solution contains 14.1 g of Cu2SO4?

Chemistry
2 answers:
balu736 [363]3 years ago
5 0
Copper(I) compounds in aqueous solutions are unstable and disproportionate:
Cu₂SO₄ = Cu + CuSO₄
Necessary for dissolution of non-aqueous solvent.

M(Cu₂SO₄)=223.16 g/mol
m(Cu₂SO₄)=14.1 g
v=0.250 L

n(Cu₂SO₄)=m(Cu₂SO₄)/M(Cu₂SO₄)

c=n(Cu₂SO₄)/v=m(Cu₂SO₄)/(vM(Cu₂SO₄))

c=14.1/(0.250*223.16)=0.253 mol/L

0.253 M

Natasha2012 [34]3 years ago
4 0

Answer:

A is 14.1 g

B is 0.0632 mol

C is 250.0 mL

D is 0.253 M

Explanation:

I divided 14.1 g (A) by 223.2 and got 0.0632 mol (B).

For the second part, I divided 0.0632 by 250.0 mL (C) since 1000 is mL.

Then I used the equation and got 0.253 M (D)

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Answer:

1. 2NaN₃(s) → 2Na(s) + 3N₂(g)

2. 14.5 g NaN₃

Explanation:

The answer is incomplete, as it is missing the required values to solve the problem. An internet search shows me these values for this question. Keep in mind that if your values are different your result will be different as well, but the solving methodology won't change.

" The airbags that protect people in car crashes are inflated by the extremely rapid decomposition of sodium azide, which produces large volumes of nitrogen gas. 1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide (NaN₃) into solid sodium and gaseous dinitrogen. 2. Suppose 71.0 L of dinitrogen gas are produced by this reaction, at a temperature of 16.0 °C and pressure of exactly 1 atm. Calculate the mass of sodium azide that must have reacted. Round your answer to 3 significant digits. "

1. The <u>reaction that takes place is</u>:

  • 2NaN₃(s) → 2Na(s) + 3N₂(g)

2. We use PV=nRT to <u>calculate the moles of N₂ that were produced</u>.

P = 1 atm

V = 71.0 L

n = ?

T = 16.0 °C ⇒ 16.0 + 273.16 = 289.16 K

  • 1 atm * 71.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 289.16 K
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Now we <u>convert N₂ moles to NaN₃ moles</u>:

  • 0.334 mol N₂ * \frac{2molNaN_{3}}{3molN_2} = 0.223 mol NaN₃

Finally we <u>convert NaN₃ moles to grams</u>, using its molar mass:

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The energy required to vaporize 155 g of butane at its boiling point: 61,723 kJ

<h3>Further explanation</h3>

Enthalpy is the amount of system heat at constant pressure.

The enthalpy is symbolized by H, while the change in enthalpy is the difference between the final enthalpy and the initial enthalpy symbolized by ΔH.

\large{\boxed{\boxed{\bold{\Delta H=H_{End}-H_{First}}}}

Delta H reaction (ΔH) is the amount of heat change between the system and its environment

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)

The standard unit is kilojoules (kJ)

The enthalpy change symbol (ΔH) is usually written behind the reaction equation.

Change in Standard Evaporation Enthalpy (ΔH vap) is a change in enthalpy at the evaporation of 1 mol liquid phase to the gas phase at its boiling point and standard pressure.

Examples of water evaporation:

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The enthalpy of evaporation is positive because its energy is needed to break the attraction between molecules in a liquid

  • 155 g of butane

relative molecular mass of butane (C₄H₁₀) = 4.12 + 10.1 = 58 gram / mol

tex]\large{\boxed{mole\:=\:\frac{grams}{relative\:molecular\:mass}}}[/tex]

\large mole\:=\:\large \frac{155}{58}

mole = 2,672

Since the heat of vaporization for butane is 23.1 kj / mol, the energy needed to evaporate 2,672 moles of butane is:

23.1 kJ / mol x 2,672 mol = 61,723 kJ

<h3>Learn more</h3>

the heat of vaporization

brainly.com/question/11475740

The latent heat of vaporization

brainly.com/question/10555500

brainly.com/question/4176497

Keywords: the heat of vaporization, butane, mole, gram, exothermic, endothermic

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