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alexandr1967 [171]
3 years ago
14

What is the answer??

Chemistry
2 answers:
Pavel [41]3 years ago
6 0

Answer: D and C

Explanation:

ANEK [815]3 years ago
4 0

free points if u no da wae

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A balloon filled with air has a volume of 0.500 L at 25.0 ºC and an internal pressure of 1.20 atm. What will the volume of the b
Alex_Xolod [135]

Answer:

0.441 L

Explanation:

At constant pressure, Using Charle's law  :-

\frac {V_1}{T_1}=\frac {V_2}{T_2}

Given ,  

V₁ = 0.500 L

V₂ = ?

T₁ = 25.0 °C

T₂ = -10.0 °C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (25 + 273.15) K = 298.15 K  

T₂ = (-10 + 273.15) K = 263.15 K  

Using above equation as:

\frac{0.500}{298.15}=\frac{V_2}{263.15}

V_2=\frac{0.500\cdot \:263.15}{298.15}

<u>New volume = 0.441 L</u>

4 0
3 years ago
A chemistry student needs 10.0g of ethanolamine for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the
iogann1982 [59]

Answer: The volume of ethanolamine the student should pour out is 9.80cm^3

Explanation:

To calculate the volume of ethanolamine we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of ethanolamine = 1.02g/cm^3

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Putting values in above equation, we get:

1.02g/cm^3=\frac{10.0g}{\text{Volume of ethanolamine}}\\\\{\text{Volume of ethanolamine}=\frac{10.0g}{1.02g/cm^3}=9.80cm^3

Thus the volume of ethanolamine the student should pour out is 9.80cm^3

4 0
4 years ago
Which would be most precise and why?
rjkz [21]

Answer:

The decimal point in each number

Explanation:

3 0
3 years ago
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Answer:

This is the same order as the definitions

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4 0
4 years ago
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why are the fields of science and mathematics so interdependent? select all that apply. scientists need to be able to illustrate
Vera_Pavlovna [14]
The correct answers among all the other choices are scientists need to be able to illustrate scientific laws and formulas, and scientists need to be able to take careful measurements. Thank you for posting your question. I hope this answer helped you. Let me know if you need more help. 
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4 years ago
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