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Annette [7]
3 years ago
12

How many liters of a 2.00 M NaOH solution are needed for a reaction that requires 3.55g of NaOH?

Chemistry
1 answer:
aleksandr82 [10.1K]3 years ago
4 0

Answer:

5 litres of solution are requried

Explanation:

Given that the typical units of

concentration

are

m

o

l

⋅

L

−

1

, to get

concentration

or

amount of moles

or

volume of solution

we just have to take the appropriate product or quotient.

We want

1.0

⋅

m

o

l

N

a

O

H

:

this the product

Volume of solution

×

Concentration

.

So we need the quotient:

Moles of solute

Concentration

=

1.0

⋅

m

o

l

0.2

⋅

m

o

l

⋅

L

−

1

=

5

⋅

L

.

(i.e.

1

1

⋅

L

−

1

=

1

⋅

L

)

Note that we go to such trouble in including the units in these calculations as an extra check on our arithmetic. Sometimes you ask yourself should I divide or should I multiply. Dimensional analysis answers our question. We wanted an answer in

litres

, and we got one. This persuades us that we did the calculation right.

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Discuss two physical and two chemical properties of Nitrogen​
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Answer:

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Explanation:

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1. The pressure of a gas is 100.0 kPa and its volume is 500.0 ml. If the volume increases to 1,000.0 ml, what is the new pressur
marta [7]

Answer:

1) The new pressure of the gas is 500 kilopascals.

2) The final volume is 1.44 liters.

3) Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

Explanation:

1) From the Equation of State for Ideal Gases we construct the following relationship:

\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}} (1)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kPa.

V_{1}, V_{2} - Initial and final pressure, measured in mililiters.

If we know that P_{1} = 100\,kPa, V_{1} = 500\,mL and V_{2} = 1000\,mL, then the new pressure of the gas is:

P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)

P_{2} = 500\,kPa

The new pressure of the gas is 500 kilopascals.

2) Let suppose that gas experiments an isothermal process. From the Equation of State for Ideal Gases we construct the following relationship:

\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}} (1)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kPa.

V_{1}, V_{2} - Initial and final pressure, measured in mililiters.

If we know that V_{1} = 3.60\,L, P_{1} = 10\,kPa and P_{2} = 25\,kPa then the new volume of the gas is:

V_{2} = V_{1}\cdot \left(\frac{P_{1}}{P_{2}} \right)

V_{2} = 1.44\,L

The final volume is 1.44 liters.

3) From the Equation of State for Ideal Gases we construct the following relationship:

\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}} (1)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kPa.

V_{1}, V_{2} - Initial and final pressure, measured in mililiters.

If we know that \frac{P_{2}}{P_{1}} = 3, then the volume ratio is:

\frac{V_{1}}{V_{2}} = 3

\frac{V_{2}}{V_{1}} = \frac{1}{3}

Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

8 0
3 years ago
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