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DaniilM [7]
3 years ago
7

The empirical formula for a compound is CH2O, and the molar mass is 180.2 g/mol. Which is the molecular formula for this compoun

d?A) C6H12O6
B) C7H16O5
C) C8H20O4
D) C3H6O3
Chemistry
1 answer:
Elodia [21]3 years ago
7 0

Answer : The correct option is, (A) C_{6}H_{12}O_6

Explanation :

Empirical formula : It is the simplest form of the chemical formula which depicts the whole number of atoms of each element present in the compound.  

Molecular formula : it is the chemical formula which depicts the actual number of atoms of each element present in the compound.  

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{molecular mass}}{\text{empirical mass}}

As we are given that the empirical formula of a compound is CH_2O and the molar mass of compound is, 180.2 gram/mol.

The empirical mass of CH_2O = 1(12) + 2(1) + 1(16) = 30 g/eq

n=\frac{\text{molecular mass}}{\text{empirical mass}}

n=\frac{180.2}{30}=6

Molecular formula = (CH_2O)_n=(CH_2O)_6=C_{6}H_{12}O_6

Thus, the molecular formula of the compound will be, C_{6}H_{12}O_6

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Answer:

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They also showed the effects of pressure on volume if temperature stayed the same is the experiment that will provide an evidence for Boyle's law.

Boyle's law states that "the volume of a fixed mass of a gas varies inversely as the pressure changes, if the temperature is constant".

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The law of conservation of mass states that mass is neither created nor destroyed. Since we have 2 g/mol of A and 3 g/mol of B then AB should be equal to the sum of their molar mass that is 

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Answer:

a=0.4\ m/s^2

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1. Given that,

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a=\dfrac{v-u}{t}\\\\a=\dfrac{6-4}{5}\\\\a=0.4\ m/s^2

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For the following reaction, 22.6 grams of nitrogen monoxide are allowed to react with 4.64 grams of hydrogen gas . nitrogen mono
antiseptic1488 [7]

Answer:

- 10.5 g of N₂

- Limiting reagent: NO

- 3.13 g of H₂ remains

Explanation:

First of all we state the reaction: 2NO(g) + 2H₂(g) → 2H₂O(l) + N₂(g)

We need to find out the limiting reactant and the excess reagent

Ratio in the reactants is 2:2. Let's convert the mass to moles:

22.6 g / 30 g/mol = 0.753 moles of NO

4.64 g / 2 g/mol = 2.32 moles of H₂

Certainly the limiting reagent is the NO and the excess reactant is the hydrogen:

- For 0.753 moles of NO, we need 0.753 moles of H₂ (we have 2.32 moles)

- For 2.32 moles of H₂, we need 2.32 moles of NO (and we don't have enough NO, because we only have 0.753 moles)

As the H₂ is the excess reagent, some moles still remains after the reaction is complete → 2.32 mol - 0.753 mol = 1.567 moles

We convert the moles to mass: 1.567 mol . 2g /1mol = 3.13 g of H₂ remains

As the NO is the limiting reagent, we can work with the equation:

We propose this rule of three: 2 moles of NO can produce 1 mol of N₂

Then, 0.753 moles of NO must produce (0.753 . 1) /2 = 0.376 moles of N₂

We convert the moles to mass 0.376 mol . 28 g / 1 mol = 10.5 g

3 0
3 years ago
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