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3 years ago

7.5 g CaCl2.9H2O (1 mol / 273.1215 g) ( 1 mol CaCl2 / 1 mol CaCl2.9H2O ) ( 110.98 g / 1 mol ) =

Chemistry

1 answer:

dezoksy [38]3 years ago

4 0

Use Socratic or mathpapa

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A piston containing 0.120moles of methane gas, CH4, has a volume of 2.12liters. If methane is added until the volume is increase

earnstyle [38]

Answer:

2.83 g

Explanation:

At constant temperature and pressure, Using Avogadro's law

$\frac {V_1}{n_1}=\frac {V_2}{n_2}$

Given ,

V₁ = 2.12 L

V₂ = 3.12 L

n₁ = 0.120 moles

n₂ = ?

Using above equation as:

$\frac{2.12}{0.120}=\frac{3.12}{n_2}$

$2.12n_2=0.12\cdot \:3.12$

$2.12n_2=0.3744$

$n_2=\frac{0.3744}{2.12}$

$n_2=0.17660$

n₂ = 0.17660 moles

Molar mass of methane gas = 16.05 g/mol

So, Mass = Moles*Molar mass = 0.17660 * 16.05 g = 2.83 g

<u>2.83 g are in the piston.</u>

8 0

4 years ago

Select the correct answer. Which substance is made of polymers? A. marble B. protein C. salt D. steel

Greeley [361]

Answer:

the answer is B proteins

Explanation:

brainliest pls

4 0

3 years ago

Read 2 more answers

Discuss three evidences that confirm that air is a mixture and not a compound

s2008m [1.1K]

Answer:

the air contains nitrogen carbon dioxide oxygen and many other gases

6 0

3 years ago

Calculate the number of different atoms in 0.2 mol ethene

muminat

Answer:

1.204 × 10²³

Explanation:

The number of atoms in a mole is always 6.022 × 10²³, known as Avogadro's number or Avogadro's constant.

To convert moles to atoms, multiply the molar amount by Avogadro's number.

(6.022 × 10²³) × 0.2

= 1.204 × 10²³

4 0

3 years ago

Cyclobutane, C4H8, consisting of molecules in which four carbon atoms form a ring, decomposes, when heated, to give ethylene. Th

frosja888 [35]

Answer:

The concentration of cyclobutane after 875 seconds is approximately 0.000961 M

Explanation:

The initial concentration of cyclobutane, C₄H₈, [A₀] = 0.00150 M

The final concentration of cyclobutane, [ $A_t$ ] = 0.00119 M

The time for the reaction, t = 455 seconds

Therefore, the Rate Law for the first order reaction is presented as follows;

$\text{ ln} \dfrac {[A_t]}{[A_0]} = \text {-k} \cdot t }$

Therefore, we get;

$k = \dfrac{\text{ ln} \dfrac {[A_t]}{[A_0]}} {-t }$

Which gives;

$k = \dfrac{\text{ ln} \dfrac {0.00119}{0.00150}} {-455} \approx 5.088 \times 10^{-4}$

k ≈ 5.088 × 10⁻⁴ s⁻¹

The concentration after 875 seconds is given as follows;

[ $A_t$ ] = [A₀]· $e^{-k \cdot t}$

Therefore;

[ $A_t$ ] = 0.00150 × $e^{5.088 \times 10^{-4} \times 875}$ = 0.000961

The concentration of cyclobutane after 875 seconds, [ $A_t$ ] ≈ 0.000961 M

6 0

3 years ago