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8090 [49]
3 years ago
5

What is the molar mass of CHCl3? 48.47 g/mol 83.92 g/mol 119.37 g/mol 121.39 g/mol

Chemistry
2 answers:
kolbaska11 [484]3 years ago
8 0

119.37 g/mol

<h3>Further explanation</h3>

Relative atomic mass (Ar) and relative molecular mass / molar mass (M)  

the molar mass/molecular mass of a compound : the sum of the relative atomic mass (Ar) of the constituent atoms

M AxBy = (x.Ar A + y. Ar B)  

Atomic weight (Ar) of :

  • C : 12.0107 g/mol  
  • H : 1.00784 g/mol  
  • Cl : 35.453 g/mol  

The molar mass of CHCl₃

\tt CHCl_3=12.0107+1,00784+(35,453\times 3)=119.37~g/mol

Jlenok [28]3 years ago
3 0

Answer:

C

Explanation:

I just got 100% on the test.

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Gaseous butane, CH3(CH2)2CH, reacts with gaseous oxygen gas, O2, to produce gaseous carbon dioxide, CO2, and gaseous water, H2O.
weeeeeb [17]

Answer:

Percentage yield of carbon dioxide is 49.9%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2CH3(CH2)2CH3 + 13O2 —> 8CO2 + 10H2O

OR

2C4H10 + 13O2 —> 8CO2 + 10H2O

Next, we shall determine the masses of butane and oxygen that reacted and the mass of carbon dioxide produced from the balanced equation. This is illustrated below:

Molar mass of butane C4H10 = (12×4) + (10×1)

= 48 + 10

= 58 g/mol

Mass of C4H10 from the balanced equation = 2 × 58 = 116 g

Molar mass of O2 = 16 × 2 = 32 g/mol

Mass of O2 from the balanced equation = 13 × 32 = 416 g

Molar mass of CO2 = 12 + (16×2)

= 12 + 32

= 44 g/mol

Mass of CO2 from the balanced equation = 8 × 44 = 352 g

Summary:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen to produce 352 g of carbon dioxide.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen.

Therefore, 34.29 g of butane will react with = (34.29 × 416) / 116 = 122.97 g of oxygen.

From the calculation made above, we can see clearly that only 122.97 g out of 165.7 g of oxygen reacted completely with 34.29 g of butane. Therefore, butane is the limiting reactant and oxygen is the excess reactant.

Next, we shall determine the theoretical yield of carbon dioxide.

In this case, we shall use the limiting reactant because it will give the maximum yield of carbon dioxide as all of it is used up in the reaction.

The limiting reactant is butane and the theoretical yield of carbon dioxide can be obtained as follow:

From the balanced equation above,

116 g of butane reacted to produce 352 g of carbon dioxide.

Therefore, 34.29 g of butane will react to produce = (34.29 × 352) / 116 = 104.05 g of carbon dioxide.

Therefore, the theoretical yield of carbon dioxide is 104.05 g

Finally, we shall determine the percentage yield of carbon dioxide as follow:

Actual yield of carbon dioxide = 51.9 g

Theoretical yield of carbon dioxide = 104.05 g

Percentage yield of carbon dioxide =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield of carbon dioxide = 51.9 / 104.05 × 100

Percentage yield of carbon dioxide = 49.9%

7 0
3 years ago
You are instructed to create 900. mL of a 0.29 M phosphate buffer with a pH of 7.8. You have phosphoric acid and the sodium salt
Aleksandr [31]

Answer:

B and C

Explanation:

When we have to do a buffer solution we always have to choose the reaction that has the <u>pKa closer to the desired pH value</u>. When we find the pKa values we will obtain:

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pKa_2=-Log[6.2x10^-^8]=7.20

pKa_3=-Log[4.8x10^-^13]=12.31

The closer value is pKa2 with a value of 7.2. Therefore we have to use the second reaction. In which  H_2PO_4^-^1 is the <u>acid</u> and HPO_4^-^2 is the <u>base</u>. Therefore the answer for the first question is B and the answer for the second question is C.

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3 years ago
How many milliliters of a 8.20 m licl solution contain 18.4 g of licl?
Dovator [93]
Your final answer is that you will need 52.9mL of the 8.20 M of LiCl

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