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3 years ago

50.0 g of NaNO3 are dissolved into enough water to make 250 mL of solution. What is the molarity of this solution?

Chemistry

1 answer:

d1i1m1o1n [39]3 years ago

7 0

Answer:

2.35 M

Explanation:

Molarity is mol/L of solution. We have to convert the g to mol and the mL to L. G to mol uses the molar mass of the compound. The molar mass of NaNO₃ is 85.00g/mol.

$50.0gNaNO3*\frac{1molNaNO3}{85.00gNaNO3} = 0.588molNaNO3$

Then you have to convert mL to L.

$250mL*\frac{1L}{1000mL} = 0.250L$

Now divide the mol by the L.

$\frac{0.588mol NaNO3}{0.250L} = 2.352 M$

Round to the smallest number of significant figures = 2.35M

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Answer:

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3 years ago

Read 2 more answers

Which statements are true of solutions?

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Explanation:

4 0

3 years ago

The molecular formula of morphine, a pain-killing narcotic, is c17h19no3. what is the molar mass? what is the mass of 6.02 x 102

kati45 [8]

Molar mass is the mass of one mole of a substance, it can be a chemical element or a compound. It is a characteristic of each pure substance. We calculate it by adding up all of the masses of the atoms involved in the compound. We calculate as follows:
atomic mass total mass
C 17 12.01 g/mol 204.17 g/mol
H 19 1.01 g/mol 19.19 g/mol
N 1 14.00 g/mol 14.00 g/mol
O 3 16.00 g/mol 48.00 g/mol
------------------------------------------------------------
Molar mass = 285.36 g/mol

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3 years ago

A buffer can be prepared by mixing two solutions. Determine if each of the following mixtures will result in a buffer solution o

GenaCL600 [577]

Answer:

The answers are in the explanation

Explanation:

A buffer is the mixture of a weak acid with its conjugate base or vice versa. Thus:

1) Mixing 100.0 mL of 0.1 M HF with 100.0 mL of 0.05 M mol KF. Will result in a buffer because HF is a weak acid and KF is its conjugate base.

2) Mixing 100.0 mL of 0.1 M NH₃ with 100.0 mL of 0.1 M NH₄Br. Will not result in a buffer because NH₃ is a strong base.

3) Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.05 M KOH. Will result in a buffer because HCN is a weak acid and its reaction with KOH will produce CN⁻ that is its conjugate base.

4) Mixing 100.0 mL of 0.1 M HCl with 100.0 mL of 0.1 M KCl Will not result in a buffer because HCl is a strong acid.

5) Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.1 M KOH Will not result in a buffer because each HCN will react with KOH producing CN⁻, that means that you will have just CN⁻ (Conjugate base) without HCN (Weak acid).

I hope it helps!

6 0

3 years ago