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Zielflug [23.3K]
3 years ago
6

What is the molecular formula of a compound with the empirical formula

Chemistry
1 answer:
ipn [44]3 years ago
5 0

Answer:  C_2H_2O_2

Explanation:

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

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

The empirical formula is CHO

The empirical weight of CHO = 1(12)+1(1)+1(16)= 29 g.

The molecular weight = 60 g/mole

Now we have to calculate the molecular formula:

n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{60}{29}=2

The molecular formula will be=2\times CHO=C_2H_2O_2

Thus molecular formula will be C_2H_2O_2

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The chemical industry is a very important contributor to the wealth of a country. For example it contributes over 1% to the Gross National Product (GNP) of European countries, which is over 6% of the total GNP produced by all manufacturing industries.
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Write the balanced equation for the reaction of acetic acid (CH3COOH) and sodium bicarbonate (NaHCO3). What type of reaction(s)
stealth61 [152]

Answer:CH3COOH + NaHCO3 > H2O + CO2(g) + CH3COONa

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Calculate the molarity (M) of a 600 ml solution of 0.54 moles of H2SO4.
icang [17]
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Hope this helps!
8 0
3 years ago
A sample containing 2.30 mol of Ne gas has an initial volume of 8.00 L. What is the final volume, in liters, when the following
marta [7]

Answer:

a. 4,00L

b. 16,00L

c. 12,31L

Explanation:

Avogadro's law says:

\frac{V_1}{n_1} =\frac{V_2}{n_2}

a. If initial conditions are 2,30mol and 8,00L and you lose one-half of atoms, that means you have 1,15mol:

\frac{8,00L}{2,30mol} =\frac{V_2}{1,15mol}

<em>V₂ = 4,00L</em>

b. If initial conditions are 2,30mol and 8,00L and you add 2,30mol, that means you have 4,60mol:

\frac{8,00L}{2,30mol} =\frac{V_2}{4,60mol}

<em>V₂ = 16,00L</em>

c. 25,0g of Ne are:

25,0g × (1mol / 20,1797g) = 1,24 moles of Ne. That means you have 2,30mol - 1,24mol = 3,54mol of Ne

\frac{8,00L}{2,30mol} =\frac{V_2}{3,54mol}

<em>V₂ = 12,31L</em>

I hope it helps!

6 0
3 years ago
How many moles is 2.55 x 10 to the power of 26 atoms of Neon?​
Yakvenalex [24]

Answer:

424 mol

Explanation:

Step 1: Given data

Number of atoms of Neon: 2.55 × 10²⁶ atoms

Step 2: Calculate the number of moles corresponding to 2.55 × 10²⁶ atoms of Neon

In order to convert atoms into moles, we need a conversion factor, which is Avogadro's number: there are 6.02 × 10²³ atoms of Neon in 1 mole of atoms of Neon.

2.55 × 10²⁶ atoms × (1 mol/6.02 × 10²³ atoms) = 424 mol

4 0
3 years ago
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