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2 years ago

What is the empirical formula for C2H4(OH)2??

Chemistry

2 answers:

k0ka [10]2 years ago

8 0

Answer : The empirical formula for $C_2H_4(OH)_2$ is, $CH_2(OH)$

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 empirical formula, we need to determine the least whole number and then divide the whole number by the least whole number.

The given molecular formula is, $C_2H_4(OH)_2$

In this the least whole number is, 2

So, the empirical formula will be, $C_{\frac{2}{2}}H_{\frac{4}{2}}(OH)_{\frac{2}{2}}=C_1H_2(OH)_1$

Hence, the empirical formula for $C_2H_4(OH)_2$ is, $CH_2(OH)$

zhannawk [14.2K]2 years ago

7 0

Empirical formula for C2H4(OH)2 is
C1H3O1

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During a titration the following data were collected. A 50.0 mL portion of an HCl solution was titrated with 0.500 M NaOH; 200.

Travka [436]

Answer: The mass of HCl present in 500 mL of acid solution is 36.5 grams

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is HCl

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?M\\V_1=50.00mL\\n_2=1\\M_2=0.500M\\V_2=200.mL$

Putting values in above equation, we get:

$1\times M_1\times 50.00=1\times 0.500\times 200\\\\M_1=\frac{1\times 0.500\times 200}{1\times 50.00}=2M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Molar mass of HCl = 36.5 g/mol

Molarity of solution = 2 M

Volume of solution = 500 mL

Putting values in above equation, we get:

$2mol/L=\frac{\text{Mass of solute}\times 1000}{36.5g/mol\times 500}\\\\\text{Mass of solute}=\frac{2\times 36.5\times 500}{1000}=36.5g$

Hence, the mass of HCl present in 500 mL of acid solution is 36.5 grams

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2 years ago

Convert 26.02 x 1023 molecules of C2H8 to grams. Round your answer to the hundredths place.

aliina [53]

Answer:

x= 138.24 g

Explanation:

We use the avogradro's number

6.023 x 10^23 molecules -> 1 mol C2H8

26.02 x 10^23 molecules -> x

x= (26.02 x 10^23 molecules * 1 mol C2H8 )/6.023 x 10^23 molecules

x= 4.32 mol C2H8

1 mol C2H8 -> 32 g

4.32 mol C2H8 -> x

x= (4.32 mol C2H8 * 32 g)/ 1 mol C2H8

x= 138.24 g

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Answer:

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