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Yuri [45]
3 years ago
12

What are the empirical formulas of a) ethylene glycol, a radiator antifreeze, molecular formula C2H6O2 b) peroxodisulfuric acid,

a compound used in bleaching agents, molecular formula H2S2O8
Chemistry
1 answer:
Temka [501]3 years ago
6 0
<h2>The answers are CH_3O and HSO_4</h2>

Explanation:

Given -

a) The molecular formula of ethylene glycol -

    C_2H_6O_2

∴ The empirical formula of ethylene glycol will be -

    CH_3O

Given -

b) The molecular formula of per-oxo-disulfuric acid (a compound used in bleaching agents) -

    H_2S_2O_8

∴ The empirical formula of per-oxo-disulfuric acid will be -

    HSO_4

Hence, the answers are CH_3O and HSO_4.

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3 years ago
If 27.3% of a sample of silver-112 decays in 1.52 hours, what is the half-life (in hours to 3 decimal places)?
ICE Princess25 [194]

<u>Answer:</u> The half life of the sample of silver-112 is 3.303 hours.

<u>Explanation:</u>

All radioactive decay processes undergoes first order reaction.

To calculate the rate constant for first order reaction, we use the integrated rate law equation for first order, which is:

k=\frac{2.303}{t}\log \frac{[A_o]}{[A]}

where,

k = rate constant = ?

t = time taken = 1.52 hrs

[A_o] = Initial concentration of reactant = 100 g

[A] = Concentration of reactant left after time 't' = [100 - 27.3] = 72.7 g

Putting values in above equation, we get:

k=\frac{2.303}{1.52hrs}\log \frac{100}{72.7}\\\\k= 0.2098hr^{-1}

To calculate the half life period of first order reaction, we use the equation:

t_{1/2}=\frac{0.693}{k}

where,

t_{1/2} = half life period of first order reaction = ?

k = rate constant = 0.2098hr^{-1}

Putting values in above equation, we get:

t_{1/2}=\frac{0.693}{0.2098hr^{-1}}\\\\t_{1/2}=3.303hrs

Hence, the half life of the sample of silver-112 is 3.303 hours.

6 0
3 years ago
The gaseous product of a reaction is collected in a 25.0L container at 27.0 C. The pressure in the container is 3.0atm and the g
NeX [460]

Answer: The molar mass of the gas is 31.6 g/mol

Explanation:

According to ideal gas equation:

PV=nRT

P = pressure of gas = 3.0 atm

V = Volume of gas = 25.0 L

n = number of moles  = ?

R = gas constant =0.0821Latm/Kmol

T =temperature =27.0^0C=(27.0+273)K=300K

n=\frac{PV}{RT}

n=\frac{3.0atm\times 25.0L}{0.0821 L atm/K mol\times 300K}=3.04moles

Moles =\frac{\text {given mass}}{\text {Molar mass}}

3.04=\frac{96.0g}{\text {Molar mass}}

{\text {Molar mass}}=31.6g/mol

The molar mass of the gas is 31.6 g/mol

4 0
3 years ago
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