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olga2289 [7]
3 years ago
5

I need some help please

Chemistry
1 answer:
Norma-Jean [14]3 years ago
5 0

Answer:

the first one to the third box

the second one to the fourth box

the third one to second box

and the fourth one to the first box

Explanation:

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When the temperature of a gas changes, it's volume decreases from 12 cm3 to 7 cm3 if the final temperature is measured to be 18°
ElenaW [278]

Answer:

The initial temperature is 499 K

Explanation:

Step 1: Data given

initial volume = 12 cm3 = 12 mL

Final volume = 7 cm3 = 7mL

The final temperature = 18 °C = 291 K

Step 2: Calculate the initial temperature

V1/T1 = V2/T2

⇒with V1 = the initial volume = 0.012 L

⇒with T1 = the initial volume = ?

⇒with V2 = the final volume 0.007 L

⇒with T2 = The final temperature = 291 K

0.012 / T1 = 0.007 / 291

0.012/T1 = 2.4055*10^-5

T1 = 0.012/2.4055*10^-5

T1 = 499 K

The initial temperature is 499 K

8 0
3 years ago
The enthalpy of vaporization for water is 40.67 kJ/mol and the enthalpy of fusion for water is
S_A_V [24]

Answer:

\Delta H_{dep}=-46.68kJ/mol

Explanation:

Hello!

In this case, since vaporization is the processes by which a liquid goes to the gas phase, fusion by which a solid goes to liquid and deposition by which a gas goes to solid; we infer that the following set up relates the enthalpies associated to each process:

\Delta H_{fus}+\Delta H_{vap}=-\Delta H_{dep}

Because deposition goes from a state with more energy to a state with less energy, therefore it is negative; in such a way, by plugging in we obtain:

\Delta H_{dep}=-(\Delta H_{fus}+\Delta H_{vap})\\\\\Delta H_{dep}=-(6.01kJ/mol+40.67kJ/mol)\\\\\Delta H_{dep}=-46.68kJ/mol

Best regards!

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