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2 years ago

.25 g of each of the following gasses are taken at 27°C and 600 mm

Chemistry

1 answer:

horsena [70]2 years ago

6 0

HI has the least volume

<h3>Further explanation</h3>

Given

Gasses and molar mass

Required

The least volume

Solution

Ideal gas law

PV=nRT

V = (nRT)/P

n = mass : MW(Molar weight=molar mass)

$\tt V=\dfrac{\dfrac{mass}{MW}RT }{P}\\\\V=\dfrac{mass.RT}{MW.P}$

so V is inversely proportional to MW(molar mass)

Because the values of P, R, T, and mass are the same, what matters is the value of the molar mass of each compound. The larger the molar mass the smaller the volume

From the available molar mass data, HI has the largest molar mass, so HI has the smallest volume

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Ch4+2o2=co2+2h2o how many moles of methane are in 20 liters of methane?

horsena [70]

Answer:

See below, please

Explanation:

1 mol CH4 ——＞2 mol H2O=2x18 g/mol= 36 g

The question is not clear. You should tell us the concentration of CH4, methane.

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Mexico City has an elevation of 7400 ft above sea level. Its normal atmospheric pressure is 589 mm Hg. At what temperature does

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Given that the volume and amount of water are kept constant,
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3 years ago

How many ions are present in iron 2 oxide

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3 years ago

Which of the following elements is a nonmetal? a. selenium (Se) b. yttrium (Y) c. barium (Ba) d. calcium (Ca) e. All of these ar

san4es73 [151]

Answer:

The correct answer option is a. Selenium (Se).

Explanation:

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3 years ago

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Caffeine, a stimulant found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 1.000 mg of caffeine

Yuliya22 [10]

Answer:

194 g/mol

Explanation:

1) Content of C:

All the C atoms in the 1.000 mg of caffeine will be found in the 1.813 mg of CO₂.

Mass of C in 1.813 mg of CO₂

Since the atomic mass of C is 12.01 g/mol and the molar of of CO₂ is 44.01, there are 12 mg of C in 44 mg of CO₂ and you can set the proporton:

12.01 mg C / 44.01 mg CO₂ = x / 1.813 g CO₂

⇒ x = 1.813 × 12.01 / 44.01 g of C = 0.49475 mg of C

Number of moles of C

number of moles = mass in g / atomic mass = 0.49475×10⁻³ g / 12.01 g/mol = 4.1195×10⁻⁵ moles = 0.041195 milimol

2) Content of H

All the H atoms in the 1.000 mg of caffeine will be found in the 0.4639 mg of H₂O

Mass of H in 0.4639 mg of H₂O

Since the atomic mass of H is 1.008 g/mol and the molar of of H₂O is 18.015 g/mol, there are 2×1.008 mg of H in 18.015 mg of H₂O and you can set the proporton:

2×1.008 mg H / 18.015 mg H₂O = x / 0.4639 mg H₂O

⇒ x = 0.4639 mg H₂O × 2 × 1.008 mg H / 18.015 mg H₂O = 0.051913 mg H

Number of moles of H

number of moles = mass in g / atomic mass = 0.051913 ×10⁻³ g / 1.008 g/mol = 5.1501× 10⁻⁵ moles = 0.051501 milimol

3) Content of N

All the N atoms in the 1.000 mg of caffeine will be found in the 0.2885 mg of N₂

Mass of N in 0.2885 mg of N₂ is 0.2885 mg

Number of moles of N

number of moles = mass in g / atomic mass = 0.2885 ×10⁻³ g / 14.007 g/mol = 2.0597× 10⁻⁵ moles = 0.020597 milimol

4) Content of O

The mass of O is calculated by difference:

Mass of O = mass of sample - mass of C - mass of H - mass of N

Mass of O = 1.000 mg - 0.49475 mg C - 0.051913 mg H - 0.2885 mg N

Mass of O = 0.1648 mg

Moles of O = 0.1648 × 10 ⁻³ g / 15.999 g/mol = 1.0303×10⁻⁵ mol = 0.01030 milimol

5) Ratios

Divide every number of mililmoles by the smallest number of milimoles:

C: 0.041195 / 0.01030 = 4
H: 0.051501 / 0.01030 = 5
N: 0.020597 / 0.01030 = 2

O: 0.01030 / 0.01030 = 1

C: 4
H: 5
N: 2
O: 1

6) Empirical formula:

C₄H₅N₂O₁

7) Calculate the approximate mass of the empirical formula:

4 × 12 + 5 × 1 + 2 × 14 + 1 × 16 = 97 g/mol

So, since that number is not between 150 and 200 g/mol, multiply by 2: 97 × 2 = 194, which is between 150 and 200.

Thus, the estimate is 194 g/mol

7 0

3 years ago