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Alecsey [184]
2 years ago
5

A 200.-milliliter sample of CO2(g) is placed in a sealed, rigid cylinder with a movable piston at 296 K and 101.3 kPa. Determine

the volume of the sample of CO2(g) if the temperature and pressure are changed to 336 K and 152.0 kPa.
Chemistry
1 answer:
bagirrra123 [75]2 years ago
4 0

Answer:

The final volume of the sample of gas V_{2} = 0.000151 m^{3}

Explanation:

Initial volume V_{1} = 200 ml = 0.0002 m^{3}

Initial temperature T_{1} = 296 K

Initial pressure P_{1} = 101.3 K pa

Final temperature T_{2} = 336 K

Final pressure P_{2} =  K pa

Relation between P , V & T is given by

P_{1} \frac{V_{1} }{T_{1} } = P_{2} \frac{V_{2} }{T_{2} }

Put all the values in the above equation we get

101.3 (\frac{0.0002}{296} )= 152 (\frac{V_{2} }{336} )

V_{2} = 0.000151 m^{3}

This is the final volume of the sample of gas.

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What is the halflife of a radioisotope if a 20-g sample becomes 10g after 16 hours
ohaa [14]

Answer:

T½ = 16hours

Explanation:

Final mass (N) = 10g

Initial mass (No) = 20g

Time (t) = 16hours

T½ = ?

T½ = In2 / λ

But λ = ?

In(N/No) = -λt

In(10/20) = -(λ * 16)

In(0.5) = -16λ

-0.693 = -16λ

λ = 0.693 / 16

λ = 0.0433

Note : λ is known as the disintegration constant

T½ = In2 / λ

T½ = 0.693 / 0.0433

T½ = 16hours

The half-life of the sample is 16hours

5 0
2 years ago
7. Two teams are competing in a tug-of-war contest. Team A is pulling at 4000N and Team B is pulling at 4900N is the opposite di
Leto [7]
Basically team B would win since it is exerting a force of 900N unlike team A ( you can tell by doing 4900N minus 4000N ). It is very unbalanced.

5 0
2 years ago
Read 2 more answers
Microwave ovens are able to cook food because they increase the_______ of water molecules in the food.
Whitepunk [10]

potential energy with the heat given to the food

4 0
2 years ago
Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown c
Alexus [3.1K]

Answer:

0.3229 M HBr(aq)

0.08436M H₂SO₄(aq)

Explanation:

<em>Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.</em>

<em />

Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).

NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)

When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.

18.45 \times 10^{-3} L NaOH.\frac{0.3500molNaOH}{1LNaOH} .\frac{1molHBr}{1molNaOH} .\frac{1}{20.00 \times 10^{-3} LHBr} =\frac{0.3229molHBr}{1LHBr} =0.3229M

<em>Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?</em>

<em />

Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).

2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)

When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.

42.18 \times 10^{-3} LNaOH.\frac{0.1000molNaOH}{1LNaOH} .\frac{1molH_{2}SO_{4}}{2molNaOH} .\frac{1}{25.00\times 10^{-3}LH_{2}SO_{4}} =\frac{0.08436molH_{2}SO_{4}}{1LH_{2}SO_{4}} =0.08436M

6 0
2 years ago
The pKs of succinic acid are 4.21 and 5.64. How many grams of monosodium succinate (FW = 140 g/mol) and disodium succinate (FW =
Varvara68 [4.7K]

Answer:

9.744g of monosodium succinate.

4.925g of disodium succinate.

Explanation:

To find pH of the buffer produced by the mixture of monosodium succinate-Disodium succinate is obtained from H-H equation:

pH = pKa + log ([Na₂Suc] / [NaHSuc])

As you want a pH of 5.28 and pKa is 5.64:

5.28 = 5.64 + log ([Na₂Suc] / [NaHSuc])

-0.36 = log ([Na₂Suc] / [NaHSuc])

0.4365 = ([Na₂Suc] / [NaHSuc]) <em>(1)</em>

<em />

As total concentration of the buffer is 100mM = 0.100M:

0.100M = [Na₂Suc] + [NaHSuc] <em>(2)</em>

Replacing (2) in (1):

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 [NaHSuc] = 0.100M - [NaHSuc]

1.4365 [NaHSuc] = 0.100M

[NaHSuc] = 0.0696M

And:

[Na₂Suc] = 0.0304M

As volume of the buffer is 1L:

[NaHSuc] = 0.0696 moles

[Na₂Suc] = 0.0304 moles

Using molar mass of both substances:

Mass of monosodium succinate:

0.0696moles * (140g / 1mol) =<em> 9.744g of monosodium succinate.</em>

Mass of disodium succinate:

0.0304moles * (162g / 1mol) =<em> 4.925g of disodium succinate.</em>

<em></em>

5 0
2 years ago
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