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Phantasy [73]
2 years ago
5

The empirical formula of a chemical substance is CH2. The molar mass of a molecule of the substance is 56.108 g/mol. What is the

molecular formula of the chemical substance?
A:C2H4
B:C4H8
C:C3H4
D:C6H6
Chemistry
2 answers:
inn [45]2 years ago
8 0

Answer:- Molecular formula of the compound is C_4H_8 .

Solution:- A molecular formula has n number of empirical formula units, where n is a whole number integer like 1,2,3 and so on.

Molar mass = n(empirical formula mass)

Empirical formula mass = 12.01+2(1.01)

= 12.01+2.02

= 14.03

Molar mass is given as 56.108. Let's plug in the values and calculate the value of n:

56.108 = n(14.03)

n=\frac{56.108}{14.03}

n = 4

Value of n is 4 means there are four empirical formula units involved to make the molecular formula of the compound.

So, the molecular formula of the compound is C_4H_8 .

ludmilkaskok [199]2 years ago
7 0

The empirical formula of a chemical substance is CH2. The molar mass of a molecule of the substance is 56.108 g/mol. What is the molecular formula of the chemical substance?

C2H4

C4H8 CORRECT ANSWER

C3H4

C6H6

I took the test and B) was correct

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2 M n O 2 + 4 K O H + O 2 + C l 2 → 2 K M n O 4 + 2 K C l + 2 H 2 O , there are 100.0 g of each reactant available. Which reacta
Sliva [168]

Answer:

The limiting reactant is KOH.

Explanation:

To find the limiting reactant we need to calculate the number of moles of each one:

\eta = \frac{m}{M}

<u>Where</u>:

η: is the number of moles

m: is the mass

M: is the molar mass

\eta_{MnO_{2}} = \frac{100.0 g}{86.9368 g/mol} = 1.15 moles  

\eta_{KOH} = \frac{100.0 g}{56.1056 g/mol} = 1.78 moles  

\eta_{O_{2}} = \frac{100.0 g}{31.998 g/mol} = 3.13 moles  

\eta_{Cl_{2}} = \frac{100.0 g}{70.9 g/mol} = 1.41 moles  

Now, we can find the limiting reactant using the stoichiometric relation between the reactants in the reaction:

\eta_{MnO_{2}} = \frac{\eta_{MnO_{2}}}{\eta_{KOH}}*\eta_{KOH} = \frac{2}{4}*1.78 moles = 0.89 moles

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\eta_{O_{2}} = \frac{\eta_{O_{2}}}{\eta_{Cl_{2}}}*\eta_{Cl_{2}} = \frac{1}{1}*1.41 moles = 1.41 moles

Similarly, we have that between O₂ and Cl₂, the limiting reactant is Cl₂.

Now, the limiting reactant between KOH and Cl₂ is:

\eta_{KOH} = \frac{\eta_{KOH}}{\eta_{Cl_{2}}}*\eta_{Cl_{2}} = \frac{4}{1}*1.41 moles = 5.64 moles

Therefore, the limiting reactant is KOH.

I hope it helps you!

6 0
2 years ago
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Answer:

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Explanation:

In the problem you are diluting the original HNO3 solution by the addition of some water. The final volume is:

290.7mL + 350.0mL = 640.7mL

And you are diluting the solution:

640.7mL / 350.0mL = 1.8306 times

As the original concentration was 12.3M, the final concentration will be:

12.3M / 1.8306 =

<h3>6.72M of HNO3</h3>
5 0
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Lets assume the gas is acting Ideally, then According to Ideal Gas Equation the density is given as,

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Where;
P  = Pressure =  1.03 atm

M  = Molar Mass =  146.06 g/mol

R  = Gas Constant =  0.08206 atm.L.mol⁻¹.K⁻¹

T  = Temperature =   297 K

Putting Values in eq. 1,

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Hint: "kilogram" means 1000 grams. There are 1,000 grams in each kilogram.

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