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kramer
3 years ago
8

Two of the simplest compounds containing just carbon and hydrogen are methane and ethane. Methane contains 0.3357 g of hydrogen

for every 1.00 g of carbon. The ratio of grams of hydrogen per gram of carbon in methane to grams of hydrogen per gram of carbon in ethane is 4:3. Determine the number of grams of hydrogen per gram of carbon in ethane.
Chemistry
1 answer:
Andreyy893 years ago
5 0
Let the ratio of grams of hydrogen per gram of carbon in methane be M, we know that:
M = 0.3357 g / 1 g

Next, lets represent the grams of hydrogen per gram of carbon in ethane be E. The final piece of information we have is:

M / E = 4/3

If we cross multiply,

3M = 4E

Now, substituting the value of M from earlier and solving for E,

E = (3 * 0.3357) / 4
E = 0.2518

There are 0.2518 grams of hydrogen per gram of carbon in ethane.
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Hydrazine reacts with oxygen according to the following equation: N2H4(g) +O2(g) → N2(g) + 2 H2O(l) How many L of N2, measured a
mihalych1998 [28]

Answer:

V ≈ 646.50 L

General Formulas and Concepts:

<u>Chemistry - Gas Laws</u>

  • Reading a Periodic Table
  • Stoichiometry
  • Combined Gas Law: PV = nRT
  • R constant - 62.4 (L · torr)/(mol · K)
  • Kelvin Conversion: K = °C + 273.15

Explanation:

<u>Step 1: Define</u>

RxN:   N₂H₄ (g) + O₂ (g) → N₂ (g) + 2H₂O (l)

Given:   34.9 °C, 755.08 torr, 914.894 g H₂O

<u>Step 2: Identify Conversions</u>

Kelvin Conversion

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Convert</u>

Stoichiometry:   914.894 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O} )(\frac{1 \ mol \ N_2}{2 \ mol \ H_2O} ) = 25.3955 mol N₂

Temperature:    34.9 + 273.15 = 308.05 K

<u>Step 4: Find Volume</u>

  1. Substitute variables:                                                                             (755.08 torr)V = (25.3955 mol)(62.4 (L · torr)/(mol · K))(308.05 K)
  2. Multiply:                                                                                                      (755.08 torr)V = 488160 L · torr
  3. Isolate <em>V</em>:                                                                                                          V = 646.502 L

<u>Step 5: Check</u>

<em>We are given 5 sig figs as our lowest. Follow sig fig rules and round.</em>

646.502 L ≈ 646.50 L

6 0
3 years ago
Be sure to answer all parts. What is the effect of each of the following on the volume of 1 mol of an ideal gas? (a) The pressur
aksik [14]

Answer

A)The volume decreases by a factor of 4

B), the volume has increased by factor of 2

Explanation:

A)Given:

P1= 760Kpa

P2 =202Kpa

The temperature changes from37C to155C

There is increase In pressure from P1 to P2

P1= 760torr.

We need to convert to Kpa

But, 1atm= 760torr

Then 760torr 101000pa

Then 101000pa = 101Kpa

We need to convert the temperature from Celsius to Kelvin

T1= 37+273= 310K

But from ideal gas, we know that PV = nRT where nR is constant

Where P= pressure

V= volume

T= temperature

n = number of moles

(P1V1/T1)=(P2V2/T2)

V1/V2 = P2/P1 * T1/T2

V1/V2 = (202/101)*(310/155)

V1/V2=4

V2= V1/4

Therefore, the volume has decreased by factor of 4

B)

Given:

P1= 2atm

P2 =101Kpa

The temperature changes from 305K to 32C

There is increase In pressure from P1 to P2

P1= 2atm

We need to convert to Kpa

But, 1atm= 760torr

Then 760torr 101000pa

Then 101000pa = 101Kpa

P1= 202.65kpa

We need to convert the temperature from Celsius to Kelvin

T2= 32+273= 305K

But from ideal gas, we know that PV = nRT

Where P= pressure

V= volume

T= temperature

n = number of moles

(P1V1/T1)=(P2V2/T2)

V1/V2 = P2/P1 * T1/T2

V1/V2 = (202/101)

V1/V2 = (101/202.65)*(305/305)

V1/V2 = 1/2

V2=2V1

Therefore, the volume has increased by factor of 2

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Number 1 is Comets and asteroids
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7 0
3 years ago
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How many mL of a 4% mass/volume Mg(NO3)2 solution would contain 1.2 grams of magnesium nitrate?
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4% mass / volume :

4 g ---------> 100 mL
1.2 g ------- ? mL

V = 1.2 * 100 / 4

V = 120 / 4

V = 30 mL

hope this helps!

7 0
3 years ago
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