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3 years ago

Methane reacts with chlorine in the presence of ultraviolet light

Chemistry

1 answer:

earnstyle [38]3 years ago

7 0

Answer:

When a mixture of methane and chlorine is exposed to ultraviolet light - typically sunlight - a substitution reaction occurs and the organic product is chloromethane. CH 4 + Cl 2 → CH 3 Cl + HCl However, the reaction doesn't stop there, and all the hydrogens in the methane can in turn be replaced by chlorine atoms.

Explanation:

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At 25°C, the equilibrium constant Kc for the reaction in thesolvent CCl4 2BrCl <----> Br2 + Cl2 is 0.141. If the initial c

ivolga24 [154]

Answer: The equilibrium concentration of bromine gas is 0.00135 M

Explanation:

We are given:

Initial concentration of chlorine gas = 0.0300 M

Initial concentration of bromine monochlorine = 0.0200 M

For the given chemical equation:

$2BrCl\rightleftharpoons Br_2+Cl_2$

Initial: 0.02 0.03

At eqllm: 0.02-2x x 0.03+x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[Br_2]\times [Cl_2]}{[BrCl]^2}$

We are given:

$K_c=0.141$

Putting values in above equation, we get:

$0.141=\frac{x\times (0.03+x)}{(0.02-2x)^2}\\\\x=-0.96,+0.00135$

Neglecting the value of x = -0.96 because, concentration cannot be negative

So, equilibrium concentration of bromine gas = x = 0.00135 M

Hence, the equilibrium concentration of bromine gas is 0.00135 M

8 0

4 years ago

NEED HELPPP PLSSSSSSSSSS

Leto [7]

I would pick the first option in the third option

4 0

3 years ago

What is the molar concentration (molarity) of 1.0 mol of KCl dissolved in 750 mL of

Nikolay [14]

Answer:

Molarity is moles per liter. You have one mole in 0.750 liters

Explanation:

8 0

3 years ago

Calculate the fraction of lattice sites that are schottky defects for cesium chloride at its melting temperature (645oc). assume

baherus [9]

Answer : 7.87 X $10^{-6}$ .

Explanation : To find the fraction of schottky defects in the given lattice of CsCl,

we use the formula, $\frac{N_{s} }{N}} = exp ( \frac{-Q_{s}}{2KT} )$

on solving with the given values , $Q_{s}= 1.86 eV$ and T as 645 + 273 K and rest are the constants.

$\frac{N_{s} }{N}} = exp ( \frac{-1.86 eV}{2 X (8.62 X 10^{-5}) X (645 +273)} )$

we get the answer as 7.87 X $10^{-6}$ .

6 0

3 years ago

Read 2 more answers

3. A compound was analyzed and found to contain 9.8 g of nitrogen, 0.70 g of

Nesterboy [21]

Answer:

HNO₃

Explanation:

Data given

Nitrogen = 9.8 g

Hydrogen = 0.70 g

Oxygen = 33.6 g

Empirical formula = ?

Solution:

Convert the masses to moles

For Nitrogen

Molar mass of N = 14 g/mol

no. of mole = mass in g / molar mass

Put value in above formula

no. of mole = 9.8 g/ 14 g/mol

no. of mole = 0.7

mole of N = 0.7 mol

For Hydrogen

Molar mass of H = 1 g/mol

no. of mole = mass in g / molar mass

Put value in above formula

no. of mole = 0.70 g/ 1 g/mol

no. of mole = 0.7

mole of H = 0.7 mol

For Oxygen

Molar mass of O = 16 g/mol

no. of mole = mass in g / molar mass

Put value in above formula

no. of mole = 33.6 g / 16 g/mol

no. of mole = 2.1

mole of O = 2.1 mol

Now we have values in moles as below

N = 0.7

H = 0.7

O = 2.1

Divide the all values on the smallest values to get whole number ratio

N = 0.7 / 0.7 = 1

H = 0.7 / 0.7 = 1

O = 2.1 / 0.7 = 3

So all have following values

N = 1

H = 1

O = 3

So the empirical formula will be HNO₃ i.e. all three atoms in simplest small ratio.

6 0

4 years ago