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barxatty [35]
3 years ago
13

To what Celsius temperature must 67.0 mL of krypton gas at 18.0°C be changed so the volume will triple? Assume the pressure and

the amount of gas are held constant. Enter your answer to four significant figures.
Chemistry
1 answer:
Mice21 [21]3 years ago
8 0

Answer: 600°C

Explanation:

This reaction is explained by Charles' law as the pressure is constant.

From the question, we obtained:

V1 = 67mL

T1 = 18°C = 18 +273 = 291K

V2 = 3V1 ( Vol is tripled) = 3x67 = 201mL

T2 =?

Applying the Charles' law,

V1 /T1 = V2 /T2

67/291 = 201 / T2

Cross multiply to express in linear form.

67xT2 = 291x201

Divide both side by 67, we have:

T2 = (291x201) /67

T2 = 873K

Converting to Celsius temperature, we have

T°C = K — 273

T°C = 873 — 273 = 600°C

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A gas has a pressure of 3.16 atm at STP. I have decided to transfer it to a container that is 3 times larger than the original v
Oliga [24]

Answer:

<u>The new pressure is 1.0533 atm</u>

<u></u>

Explanation:

According  to<u> Boyle's Law :</u> The Pressure of fixed amount of gas is inversely proportional to Volume at constant temperature.

PV = Constant

P1V1 = P2V2

P_{1}V_{1}=P_{2}V_{2}.....(1)

P1 = 3.16 atm

Accprding to question ,

V1 = V

V2 = 3 V

Insert the value of V1 , V2 and P1 in the equation(1)

P_{1}V_{1}=P_{2}V_{2}

3.16\times V=P_{2}\times 3V

V and V cancel each other

3.16=P_{2}\times 3

P_{2}=\frac{3.16}{3}

P_{2}=1.05atm

5 0
3 years ago
A science club made pine wood cars. Each car was set on the same track and then released. The distance traveled was measured. Li
Minchanka [31]
Isn't this a math problem?
If it is the the answer should be 102.

10 decimeters=1 meter

27x10=270
270-168=102
7 0
3 years ago
Solve the following equation (y = 1.2345x – 0.6789) for x, given that y = 0.570
Andrews [41]

x = 1.01

Explanation:

Given equation:

   y = 1.2345x – 0.6789

   y = 0.570

Problem:

Solving for x

The variables in this equation are y and x

They can take up any value since they are variables.

  Since we have been given y = 0.570

        y = 1.2345x – 0.6789

To solve for x, we simply substitute for y in the equation;

     since  y = 0.570

     0.57 = 1.2345x – 0.6789

    add 0.6789 to both sides;

 0.57 + 0.6789 =  1.2345x – 0.6789 + 0.6789

  1.2489‬ =   1.2345x

Divide both sides by  1.2345

 \frac{1.2489}{  1.2345} = \frac{  1.2345x }{ 1.2345}

    x = 1.01

learn more:

Equations brainly.com/question/9045597

#learnwithBrainly

 

7 0
3 years ago
Rank these acids according to their expected pKa values.
givi [52]

Answer:

According to their expected pKa values, the order of those acids should be:

1- Cl2CHCOOH is the strongest acid and the lowest pKa.

2- ClCH2COOH is a strong acid, but no more than the first. Medium pKa value.

3- ClCH2CH2COOH is a strong acid, but no more than the two previous acids. High pKa value.

4- CH3CH2COOH  is the weakest acid, so the highest pKa value.

Explanation:

The pKa values are the negative logarithm of dissociation constant. It represents the relative strengths of the acids. Stronger acids show smaller pKa values and weak acids present larger pKa value. The stronger the acid, the weaker it's the conjugate base. The larger the pKa of the conjugate base, the stronger the acid. The strength of an acid is inversely related to the strength of its conjugate.

Conjugate bases are the substance that has one less proton than the parent acid. The conjugate base of the acid presented in the problem are:

ClCH2COOH -> ClCH2COO-  + H+

ClCH2CH2COOH -> ClCH2CH2COO- + H+

CH3CH2COOH -> CH3CH2COO- + H+

Cl2CHCOOH -> Cl2CHCOO - + H+

Cl2CHCOOH. The negative charge presented on its conjugate base is by resonance and inductive effect. This is the strongest acid.

ClCH2COOH. A negative charge is stabilized by resonance and electron-withdrawing but only one atom is present. So this acid is less strong than the first one.

ClCH2CH2COOH. The negative charge is stabilized by resonance and electron-withdrawing atom but the effect is less compared to the two acids showed previously.

CH3CH2COOH. The negative charge is stabilized by resonance and destabilized due to CH3 group. This is the weakest acid among the problem.

Stronger acids have smaller pKa values and weak acids have larger pKa values. Due to the information present in this problem, Cl2CHCOOH is the strongest acid and the lowest pKa. CH3CH2COOH is the weakest acid, so the highest pKa value.

Finally, we can conclude that according to their expected pKa values, the order of those acids should be:

1- Cl2CHCOOH is the strongest acid and the lowest pKa.

2- ClCH2COOH is a strong acid, but no more than the first. Medium pKa value.

3- ClCH2CH2COOH is a strong acid, but no more than the two previous acids. High pKa value.

4- CH3CH2COOH  is the weakest acid, so the highest pKa value.

3 0
3 years ago
Explain how transverse waves can be produced on a rope. Then describe how pieces of the rope move as waves pass.
andrew-mc [135]
Transverse waves can be produced on a rope by moving one end of the rope up and down.The movement causes motion in the particles that make up the rope and the rope itself becomes the medium. The particles move perpendicular to the propagation. The movement also causes crests(highest point of the wave) and troughs (lowest point of the wave) which move along the direction of propagation.
6 0
3 years ago
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