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3 years ago

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The volume of a gas-filled balloon is 30.0 L at 313 K and 1520 mm Hg pressure. What would the volume be at standard temperature

and pressure (STP)?

1 answer:

pashok25 [27]3 years ago

4 0

313 k
step-by-step-explanation

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A chemical change involves a change in ________. A chemical change involves a change in ________. phase the elements of a molecu

Firlakuza [10]

Answer:

All of the above.

Explanation:

Chemical changes occur when a substance combines with another to form a new substance, called chemical synthesis or, alternatively, chemical decomposition into two or more different substances.

This process is not reversible and a change of energy that is sometimes heat is given off.

4 0

3 years ago

8,000 meters into miles (show work)

pickupchik [31]

Answer:

8000 metres is equal to 4.971 miles

equation

8000 (number of meters) divided by 1609.34 ( equivalent of miles in one meter)

is equal to 4.971

7 0

3 years ago

Read 2 more answers

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]

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

<u>Explanation:</u>

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)

<u>For NaOH:</u>

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:

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

<u>For sulfuric acid:</u>

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

<u>For sodium sulfate:</u>

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:

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

<u>For sulfuric acid:</u>

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:

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

<u>For NaOH:</u>

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

In the reaction:

Ghella [55]

Answer:

The answer is D because NH4+and OH- it will be

N-1. N-1,

H-5

O-1 H-5

O-1

5 0

3 years ago

Calculate the silver ion concentration in a saturated solution of silver(i) sulfate (ksp = 1.4 × 10–5).

deff fn [24]

Answer:

= 0.030 M

Explanation:

We can take x to be the concentration in mol/L of Ag2SO4 that dissolves

Therefore; concentration of Ag+ is 2x mol/L and that of SO4^2- x mol/L.

Ksp = 1.4 x 10^-5

Ksp = [Ag+]^2 [SO42-]

= (2x)^2(x)

= 4x^3

Thus;

4x^3 = 1.4 x 10^-5

= 0.015 M

molar solubility = 0.015 M

But;

[Ag+]= 2x

Hence; silver ion concentration is

= 2 x 0.015 M

= 0.030 M

6 0

3 years ago