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2 years ago

9

State your claim. Make sure your claim fully explains how you can distinguish between materials that appear similar. HELPPPPPP M

EEEEE PLZZZZZZZZ!!!!!!!!!!!!!!!!!!!!!

2 answers:

snow_lady [41]2 years ago

7 0

The right one is darker than the left

asambeis [7]2 years ago

5 0

The material on the right is a darker shade than the one on the left. Also the one on the right also is less of a block and a little more shagged

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3 years ago

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The vapor pressure of water is 1.00 atm at 373 K, and the enthalpy of vaporization is 40.68 kJ mol!. Estimate the vapor pressure

Yuki888 [10]

Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate $\Delta H_{vap}$ of the reaction, we use clausius claypron equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = vapor pressure at temperature $T_1$

$P_2$ = vapor pressure at temperature $T_2$

$\Delta H_{vap}$ = Enthalpy of vaporization

R = Gas constant = 8.314 J/mol K

1) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =363 K

$T_2$ = final temperature =373 K

$P_2=1 atm, P_1=?$

Putting values in above equation, we get:

$\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]$

$P_1=0.69671 atm \approx 0.6970 atm$

The vapor pressure at temperature 363 K is 0.6970 atm

2) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =373 K

$T_2$ = final temperature =383 K

$P_1=1 atm, P_2?$

Putting values in above equation, we get:

$\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]$

$P_2=1.4084 atm \approx 1.410 atm$

The vapor pressure at 383 K is 1.410 atm

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3 years ago

Which of the following terms describes an accumulation of rocky, sandy, or clayey material deposited at the end of a glacier?

OLEGan [10]

Answer:

Option (D)

Explanation:

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Thus, the correct answer is option (D).

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