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2 years ago

a chemistry teacher adds 50.0 ml of 1.50 m h2so4 solution to 200 ml of water. What is the concentration of the final solution?

Chemistry

1 answer:

nevsk [136]2 years ago

5 0

Answer:0.300M

Explanation:1) Data:

a) Initial solution

M = 1.50M

V = 50.0 ml = 0.050 l

b) Solvent added = 200 ml = 0.200 l

2) Formula:

Molarity: M = moles of solute / volume of solution is liters

3) Solution:

a) initial solution:

Clearing moles from the molarity formula: moles = M × V

moles of H₂SO₄ = M × V = 1.5M × 0.050 l = 0.075 mol

b) final solution:

i) Volumen of solution = 0.050 l + 0.200l = 0.250l

ii) M = 0.075 mol / 0.250 l = 0.300M ← answeer

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3 years ago

Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe

leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as molarity, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

$c = \frac{n}{V}$

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V = volume (unit L)

Find the molar mass (M) of potassium hydroxide.

$M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}$

$M_{KOH} = 56.106 \frac{g}{mol}$

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$n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}$

$n_{KOH} = 0.25941(9)mol$

Carry one insignificant figure (shown in brackets).

Convert the volume of water to litres.

$V = \frac{500.0mL}{1}*\frac{1L}{1000mL}$

$V = 0.5000L$

Here, carrying an insignificant figure doesn't change the value.

Calculate the concentration.

$c = \frac{n}{V}$

$c = \frac{0.25941(9)mol}{0.5000 L}$

$c = 0.5188(3) \frac{mol}{L}$ <= Keep an insignificant figure for rounding

$c = 0.5188 \frac{mol}{L}$ <= Rounded up

$c = 0.5188M$ <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

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3 years ago

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2 years ago

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exis [7]

Answer is: 127 grams rams of metallic copper can be obtained.
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n(Al) = m(Al) ÷ M(Al).
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n(Cu) = 2 mol.
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