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Dominik [7]
3 years ago
5

1. A sample of oxygen is collected over water at 22 ° C and 762 torr. What is the partial pressure of the dry oxygen? The vapor

pressure of water at 22°C is 19.8 torr. A. 742 torr B. 782 torr C. 784 torr D. 750. Torr
Chemistry
1 answer:
Rom4ik [11]3 years ago
8 0

Answer: The partial pressure of the dry oxygen is 742 torr

Explanation:

Dalton's Law of Partial Pressure states that the total pressure exerted by a mixture of gases is the sum of partial pressure of each individual gas present. Thus P(total)=P_1+P_2 .........

Given; Total pressure = 762 torr

partial pressure of water = 19.8 torr

partial pressure of dry oxygen = ? torr

Total pressure  = partial pressure of water + partial pressure of dry oxygen

762 torr = 19.8 torr = partial pressure of dry oxygen

partial pressure of dry oxygen = 742 torr

The partial pressure of the dry oxygen is 742 torr

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2.A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a s
Aleks04 [339]

Answer:

In order to prepare 10 mL, 5 μM; <em> 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.</em>

In order to prepare 10 mL, 10 μM; <em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM; <em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM; <em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

Molarity x volume (initial)= Molarity x volume (final).

In order to prepare 10 mL, 5 μM from 25 μM solution,

Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?

25 x initial volume = 5 x 10

Initial volume = 50/25

                       = 2 mL

<em>2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

<em />

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

<em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

<em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

<em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

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3 years ago
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D. One Month Interval
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A crate holds 20 boxes of sewing supplies. Each box contains 8 spools of thread. Each spool has 250 meters of thread wrapped aro
exis [7]

Answer: 40000

Explanation:

250*8=2,000

2000*20=40000

5 0
2 years ago
An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.
Tju [1.3M]

Answer:

a) \Delta S

b) entropy of the sistem equal to a), entropy of the universe grater than a).

Explanation:

a) The change of entropy for a reversible process:

\delta S=\frac{\delta Q}{T}

\Delta S=\frac{Q}{T}

The energy balance:

\delta U=[tex]\delta Q- \delta W

If the process is isothermical the U doesn't change:

0=[tex]\delta Q- \delta W

\delta Q= \delta W

Q= W

The work:

W=\int_{V1}^{V2}P*dV

If it is an ideal gas:

P=\frac{n*R*T}{V}

W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV

Solving:

W=n*R*T*ln(V2/V1)

Replacing:

\Delta S=\frac{n*R*T*ln(V2/V1)}{T}

\Delta S=n*R*ln(V2/V1)}

Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

\Delta S

b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.

7 0
3 years ago
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