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dusya [7]
3 years ago
15

A solution of 10 M NaOH was used to prepare 2 L of 0.5 M NaOH. How many mL of the original NaOH solution are needed?

Chemistry
1 answer:
boyakko [2]3 years ago
7 0
The equation to solve problem is (M1) (V1) = (M2) (V2)
M1= 10M, V1= original NaOH solution, M2= 0.5 M, V2= 2L
First, the answer needs to be in ml., so convert 2L to ml = 2000ml
(10M) (V1) = (0.5M) (2000ml)
(10M)(V1) = 1000 M/ml
V1 = 1000M/ml / 10M, the Mmoles cancel out leaving just ml.
V1 = 100ml, which is the original NaOH solution needed.
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100.mL of a .795 M solution of KBr is diluted to 500.mL. what is the new concentration of the solution?
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Answer:

0.159 M

Explanation:

convert from mL to L then use the equation:

M1V1 = M2V2

rearrange to find M2

\frac{M1V1}{V2} = M2

\frac{(0.795 M)(0.100 L)}{0.500 L} = 0.159 M

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What is the balanced form of the chemical equation shown below?
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Explanation:

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How much heat is absorbed/released when 20.00 g of NH3(g) reacts in the presence of excess O2 (g) to produce NO (g) and H2O (l)
FrozenT [24]

Answer:

a. 342.9 kJ of heat are absorbed.

Explanation:

Calculation of the moles of NH_3 as:-

Mass = 20.00 g

Molar mass of NH_3 = 17.031 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{20.00\ g}{17.031\ g/mol}

Moles= 1.1743\ mol

Given that:- \Delta H=+1168\ kJ

It means that 1 mole of NH_3 undergoes reaction and absorbs 1168\ kJ of heat

So,

1168 mole of NH_3 undergoes reaction and absorbs 1168\times 1168\ kJ of heat

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7 0
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What is the change of state from solid to liquid called?
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What is the vapor pressure at 20 °c of an ideal solution prepared by the addition of 7.38 g of the nonvolatile solute urea, co(n
Romashka [77]

Answer:

83.24 mmHg.

Explanation:

  • <em>The vapor pressure of the solution (Psolution) = (Xmethanol)(P°methanol).</em>

where, Psolution is the vapor pressure of the solution,

Xmethanol is the mole fraction of methanol,

P°methanol is the pure vapor pressure of methanol.

  • We need to calculate the mole fraction of methanol (Xmethanol).

<em>Xmethanol = (n)methanol/(n) total.</em>

where, n methanol is the no. of moles of methanol.

n total is the total no. of moles of methanol and urea.

  • We can calculate the no. of moles of both methanol and urea using the relation: n = mass/molar mass.

n of methanol = mass/molar mass = (56.9 g)/(32.04 g/mol) = 1.776 mol.

n of urea = mass/molar mass = (7.38 g )/(60.06 g/mol) = 0.123 mol.

∴ Xmethanol = (n)methanol/(n) total = (1.776 mol)/(1.776 mol + 0.123 mol) = 0.935.

<em>∴ Psolution = (Xmethanol)(P°methanol)</em> = (0.935)(89.0 mmHg) =<em> 83.24 mmHg.</em>

7 0
3 years ago
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