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denpristay [2]
2 years ago
11

What does your experiment suggest about what happens when icebergs melt into the ocean? Based on your data, do you think the sea

-surface temperatures are warmer or cooler due to the melting water? Make sure you explain how your experiment helps you answer this question.
Chemistry
1 answer:
11Alexandr11 [23.1K]2 years ago
8 0
What experiment is it referring to? What type of experiment did you do? I can't answer the question, unless you share your observations and data from the experiment. =/ 
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A sample of damp air in a 1.00 L container exerts a total pressure of 741.0 torr at 20 oC; but when it is cooled to -10 oC, the
AlexFokin [52]

From the information presented in the question, the number of molecules present of water present is obtained 2.41 × 10^21 molecules.

From the information we have;

Volume of the damp air =  1 L

Pressure of the damp air =  741.0 torr or 0.975 atm

Temperature of the gas = 20 oC + 273 = 293 K

R = 0.082 atm LK-1mol-1

Number of moles = ?

n =PV/RT

n = 0.975 × 1/0.082 ×  293

n = 0.041 moles

Volume of water vapor = 1 L

Temperature of water = -10 oC + 273 = 263 K

Pressure of the gas = 607.1 torr or 0.799 atm

R = 0.082 atm LK-1mol-1

n= PV/RT

n = 0.799 × 1/ 0.082 × 263

n = 0.037 moles

Number of moles of water = 0.041 moles -  0.037 moles = 0.004 moles

If 1 mole = 6.02 × 10^23 molecules

0.004 moles = 0.004 moles × 6.02 × 10^23 molecules/1 mole

= 2.41 × 10^21 molecules

Learn more: brainly.com/question/2510654

5 0
2 years ago
What is the boiling point of water at sea level ??​
TEA [102]

Answer:

212 °F

Explanation:

4 0
3 years ago
Read 2 more answers
Find the number of kilograms in 15.40 moles of C5H10O5
OLga [1]
C5H10O5 would weight a total of 2.312 kilograms.
5 0
3 years ago
25.0 mL of a hydrofluoric acid solution of unknown concentration is titrated with 0.200 M NaOH. After 20.0 mL of the base soluti
lesantik [10]

Answer:

[HF]₀ = 0.125M

Explanation:

NaOH + HF => NaF + H₂O

Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3.   This is 0.089M NaF and 0.001M HF remaining.

=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.

                HF  ⇄    H⁺    +      F⁻

C(eq)       [HF]     10⁻³M      0.089M (<= soln after adding 20ml 0.200M NaOH)

Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka

[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M

6 0
3 years ago
Which graph correctly shows the effect on the freezing point caused by increasing the molality of a solution ?
beks73 [17]
The answer to your question is graph A :)
6 0
3 years ago
Read 2 more answers
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