What is the noble-gas electron configuration for bromine?

A fire extinguisher has a pressure of 3 atm at 25 degrees C. What is the final pressure, in atmospheres, when the fire extinguis

posledela

Answer:

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Explanation:

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5 0

3 years ago

A student reacts 20.0 mL of 0.248 M H2SO4 with 15.0 mL of 0.195 M NaOH. Write a balanced chemical equation to show this reaction

Molodets [167]

Answer: The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For NaOH:

Initial molarity of NaOH solution = 0.195 M

Volume of solution = 15.0 mL = 0.015 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.195M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.195mol/L\times 0.015L)=2.925\times 10^{-3}mol$

For sulfuric acid:

Initial molarity of sulfuric acid solution = 0.248 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.248M=\frac{\text{Moles of }H_2SO_4}{0.020L}\\\\\text{Moles of }H_2SO_4=(0.248mol/L\times 0.020L)=4.96\times 10^{-3}mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, $2.925\times 10^{-3}$ moles of KOH will react with = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sulfuric acid

As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, NaOH is considered as a limiting reagent because it limits the formation of product.

Excess moles of sulfuric acid = $(4.96-1.462)\times 10^{-3}=3.498\times 10^{-3}mol$

By Stoichiometry of the reaction:

2 moles of KOH produces 1 mole of sodium sulfate

So, $2.925\times 10^{-3}$ moles of KOH will produce = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sodium sulfate

For sodium sulfate:

Moles of sodium sulfate = $1.462\times 10^{-3}moles$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sodium sulfate}=\frac{1.462\times 10^{-3}}{0.035}=0.0418M$

For sulfuric acid:

Moles of excess sulfuric acid = $3.498\times 10^{-3}mol$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sulfuric acid}=\frac{3.498\times 10^{-3}}{0.035}=0.0999M$

For NaOH:

Moles of NaOH remained = 0 moles

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of NaOH}=\frac{0}{0.050}=0M$

Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

8 0

3 years ago

Do elements of the halogen family lose one electron to become halide ions

lana66690 [7]

Its on Google..."The group number of a representative element in the periodic table is related to the number of valence electrons it has. ... Elements of the halogen family lose one electron to become halide ions."

5 0

3 years ago

A gas has an initial pressure of 120 kPa, a volume of 23 liters, and a temperature of 20°C. If the pressure is raised to 150 kPa

ladessa [460]

Answer:

virgil Cassidy n.v

Christian doctorbbn nnn

Explanation:

Verizon mlm

7 0

2 years ago

Can someone please help I need help?!?!

anzhelika [568]

Answer:

a: 6 moles

b: 6.75 moles

c: 5 grams

Explanation:

a: mole ratio 2:3

b: mole ratio 2:3

c: mole ratio 2:2

7 0

3 years ago