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kobusy [5.1K]
2 years ago
8

Determine the number of moles of C in each sample

Chemistry
1 answer:
stepan [7]2 years ago
7 0

<u>Explanation:</u>

  • <u>In </u>CH_4<u>:</u>

Given moles = 2.0 moles

1 mole of methane contains 1 mole of carbon and 4 moles of hydrogen

Moles of carbon in CH_4=(2.0\times 1)=2.0 moles

  • <u>In </u>C_2H_6<u>:</u>

Given moles = 0.175 moles

1 mole of ethane contains 2 moles of carbon and 6 moles of hydrogen

Moles of carbon in C_2H_6=(0.175\times 2)=0.35 moles

  • <u>In </u>C_4H_{10}<u>:</u>

Given moles = 4.21 moles

1 mole of butane contains 4 moles of carbon and 10 moles of hydrogen

Moles of carbon in C_4H_{10}=(4.21\times 4)=16.84 moles

  • <u>In </u>C_8H_{18}<u>:</u>

Given moles = 24.5 moles

1 mole of octane contains 8 moles of carbon and 18 moles of hydrogen

Moles of carbon in C_8H_{18}=(24.5\times 8)=196 moles

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Heat never travels from a cooler object to a warmer object.
prohojiy [21]

Answer:

I would say False.

4 0
2 years ago
The chemical equation, 2 Cr + 3 Fe(NO3)2 - 3 Fe + 2 Cr(NO3)3, is an
lana66690 [7]

Answer:

Single replacement reaction (aka single displacement reaction)

Explanation:

In a single replacement reaction, one element is substituted for another in a compound to create a new compound and a new element in the products. The general form is:

A + BC --> B + AC

In the case of this question, Cr and Fe "trade places."

6 0
2 years ago
Using the van der Waals equation, the pressure in a 22.4 L vessel containing 1.00 mol of neon gas at 100 °C is ________ atm. (a
svetoff [14.1K]

Answer:

The answer to your question is        P = 1.357 atm

Explanation:

Data

Volume = 22.4 L

1 mol

temperature = 100°C

a = 0.211 L² atm

b = 0.0171 L/mol

R = 0.082 atmL/mol°K

Convert temperature to °K

Temperature = 100 + 273

                      = 373°K

Formula

               (P + \frac{a}{v^{2}} )(v - b) = RT

Substitution

               (P + \frac{0.211}{22.4})(22.4 - 0.0171) = (0.082)(373)

Simplify

               (P + 0.0094)(22.3829) = 30.586

Solve for P

                           P + 0.0094 = \frac{30.586}{22.3829}

                           P + 0.0094 = 1.366

                                 P = 1.336 - 0.0094

                                P = 1.357 atm

7 0
3 years ago
What is the mass of a sample of metal that is heated from 58.8°C to 88.9°C with a
Vadim26 [7]

Answer:

\boxed {\boxed {\sf 333 \ grams}}

Explanation:

We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

Q= mc \Delta T

The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.

  • ΔT= final temperature - initial temperature

The sample was heated <em>from </em> 58.8 degrees Celsius to 88.9 degrees Celsius.

  • ΔT= 88.9 °C - 58.8 °C = 30.1 °C

Now we know three variables:

  • Q= 4500.0 J
  • c= 0.4494 J/g°C
  • ΔT = 30.1 °C

Substitute these values into the formula.

4500.0 \ J = m (0.4494 \ J/g \textdegree C)(30.1 \textdegree C)

Multiply on the right side of the equation. The units of degrees Celsius cancel.

4500.0 \ J = m (13.52694 J/g)

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}

The units of Joules cancel.

\frac {4500.0 \ J }{13.52694 J/g}= m

332.6694729 \ g =m

The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

333 \ g \approx m

The mass of the sample of metal is approximately <u>333 grams.</u>

8 0
2 years ago
Need help balancing these ​
kari74 [83]
The answer would be 1,3,1,3
3 0
2 years ago
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