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3 years ago

What is the molarity of a solution containing 11.2 g Ca(N03)2 in 175 mL of solution?

Chemistry

1 answer:

lisov135 [29]3 years ago

6 0

Answer:

11.2 g Can(N03)2 in 175 mL of solution?

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Help me clarify please!

zaharov [31]

Answer:

Round to the number of significant figures in the original question. However, if you're going to proceed with further calculations using this mass, it's best not to round, as rounding will cause your answer to be less precise.

Explanation:

7 0

2 years ago

Is this right? giving brainlt if you actually checked

Maksim231197 [3]

Answer:

yes it is

Explanation:

5 0

2 years ago

Read 2 more answers

Write the balanced chemical equation between H2SO4H2SO4 and KOHKOH in aqueous solution. This is called a neutralization reaction

const2013 [10]

Answer:

0.185M sulfuric acid

Explanation:

Based on the reaction:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

1 mole of sulfuric acid reacts with 2 moles of KOH

Initial moles of H₂SO₄ and KOH are:

H₂SO₄: 0.750L ₓ (0.470mol / L) = 0.3525 moles of H₂SO₄

KOH: 0.700L ₓ (0.240mol / L) = 0.168 moles of KOH

The moles of sulfuric acis that react with KOH are:

0.168mol KOH ₓ (1 mole H₂SO₄ / 2 moles KOH) = 0.0840 moles of sulfuric acid.

Thus, moles that remain are:

0.3525moles - 0.0840 moles = 0.2685 moles of sulfuric acid remains

As total volume is 0.700L + 0.750L = 1.450L, concentration is:

0.2685mol / 1.450L = 0.185M sulfuric acid

8 0

3 years ago

Hey what's 10 & 11 y'all im dumb

lesya [120]

Answer:

10 is 2.89g/ml

11 in attachment

Explanation:

10. D=M/V

M=250G-120G=130G

V=75ML-30ML=45ML

D=130/45=2.89g/ml

3 0

3 years ago

The enthalpy of fusion of solid n-butane is 4.66 kJ/mol. Calculate the energy required to melt 58.3 g of solid n-butane.

adelina 88 [10]

Answer : The energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

Explanation :

First we have to calculate the moles of n-butane.

$\text{Moles of n-butane}=\frac{\text{Mass of n-butane}}{\text{Molar mass of n-butane}}$

Given:

Molar mass of n-butane = 58.12 g/mole

Mass of n-butane = 58.3 g

Now put all the given values in the above expression, we get:

$\text{Moles of n-butane}=\frac{58.3g}{58.12g/mol}=1.00mol$

Now we have to calculate the energy required.

$Q=\frac{\Delta H}{n}$

where,

Q = energy required

$\Delta H$ = enthalpy of fusion of solid n-butane = 4.66 kJ/mol

n = moles = 1.00 mol

Now put all the given values in the above expression, we get:

$Q=\frac{4.66kJ/mol}{1.00mol}=4.66kJ$

Thus, the energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

7 0

3 years ago