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Ira Lisetskai [31]
3 years ago
9

Pls help me. Show proof on why it’s that answer.

Chemistry
1 answer:
djverab [1.8K]3 years ago
7 0
Prevent additional sources of pollution because if the water gets polluted it can not be used and/or it won’t be pure/clean anymore.
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Which statement describes how technology has increased our information on Mars?
Luden [163]

The Curiosity rover found sulfur compounds in rocks and carbon in organic compounds like propane, butene, benzene, toluene and thiophene.

It detected methane, not in soil samples, but in the Martian atmosphere.

It did not discover helium in underground pockets. The hole it can drill is only 5 cm deep.

4 0
3 years ago
Read 2 more answers
What would be the freezing point of a solution that has a molality of 1.324 m which was prepared by dissolving biphenyl (C12H10)
lbvjy [14]

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

A solution is prepared by dissolving biphenyl into naphthalene. We can calculate the freezing point depression (ΔT) for naphthalene using the following expression.

\Delta T = i \times Kf \times m =   1 \times 6.90 \°C/m  \times 1.324m = 9.14  \°C

where,

  • i: van 't Hoff factor (1 for non-electrolytes)
  • Kf: cryoscopic constant
  • m: molality

The normal freezing point of naphthalene is 80.26 °C. The freezing point of the solution is:

T = 80.26 \° C - 9.14 \° C = 71.12 \° C

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

Learn more: brainly.com/question/2292439

3 0
2 years ago
Why aluminum in 1820 had more value than gold​
bonufazy [111]

Answer:

Aluminum had more value than gold in the 1820s, because it was harder to obtain. Hope this helped :)

4 0
2 years ago
A sample of xenon gas occupies a volume of 6.80 L at 52.0°C and 1.05 atm. If it is desired to increase the volume of the gas sam
Pavlova-9 [17]

Answer:

207.03°C

Explanation:

The following data were obtained from the question:

V1 (initial volume) = 6.80 L

T1 (initial temperature) = 52.0°C = 52 + 273 = 325K

P1 (initial pressure) = 1.05 atm

V2 (final volume) = 7.87 L

P2 (final pressure) = 1.34 atm

T2(final temperature) =?

Using the general gas equation P1V1/T1 = P2V2/T2, the final temperature of the gas sample can be obtained as follow:

P1V1/T1 = P2V2/T2

1.05 x 6.8/325 = 1.34 x 7.87/T2

Cross multiply to express in linear form as shown below:

1.05 x 6.8 x T2 = 325 x 1.34 x 7.87

Divide both side by 1.05 x 6.8

T2 = (325 x 1.34 x 7.87) /(1.05 x 6.8)

T2 = 480.03K

Now, let us convert 480.03K to a number in celsius scale. This is illustrated below:

°C = K - 273

°C = 480.03 - 273

°C = 207.03°C

Therefore, the final temperature of the gas will be 207.03°C

5 0
3 years ago
In this reaction what is the correct coefficient for sodium chloride? pb(no3)2 ? nacl → pbcl2 ? nano3
Dovator [93]
<span>pb(no3)2 + 2 nacl → pbcl2 + 2nano3</span>
4 0
3 years ago
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