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yuradex [85]
3 years ago
7

When the temperature of a gas changes, it's volume decreases from 12 cm3 to 7 cm3 if the final temperature is measured to be 18°

C what is the initial temperature in Kelvin units?
pls help :)​
Chemistry
1 answer:
ElenaW [278]3 years ago
8 0

Answer:

The initial temperature is 499 K

Explanation:

Step 1: Data given

initial volume = 12 cm3 = 12 mL

Final volume = 7 cm3 = 7mL

The final temperature = 18 °C = 291 K

Step 2: Calculate the initial temperature

V1/T1 = V2/T2

⇒with V1 = the initial volume = 0.012 L

⇒with T1 = the initial volume = ?

⇒with V2 = the final volume 0.007 L

⇒with T2 = The final temperature = 291 K

0.012 / T1 = 0.007 / 291

0.012/T1 = 2.4055*10^-5

T1 = 0.012/2.4055*10^-5

T1 = 499 K

The initial temperature is 499 K

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A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0
2 years ago
When 8.0 moles of methanol are reacted with 9.0 moles of oxygen, how many moles of water vapor does the reaction produce?
nadya68 [22]

Answer:

12 moles of H₂O are formed in this combustion.

Explanation:

First of all, think the reaction:

2CH₃OH (l) +  3O₂ (g) →  2CO₂ (g) + 4H₂O (g)

Ratio in the reactants is 2:3, so 2 mol of methanol need 3 mol of oxygen to react. Then 8 mol of CH₃OH, will need (8.3)/2 = 12 moles of O₂

We have 9 moles of O₂, so this is the limiting reactant.

3 mol of oxygen produce 4 mol of water

Then, 9 mol of oxygen will produce ( 9 .4)/3 = 12 moles

3 0
3 years ago
If 12.1 kilograms of al2o3(s), 60.4 kilograms of naoh(l), and 60.4 kilograms of hf(g) react completely, how many kilograms of cr
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m(Na₃AlF₆) = 37,28 mol · 209,94 g/mol.
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