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3 years ago

12

A 50 g piece of aluminum is heated to 800c and then dropped into cold water, where the metal's temperature drops to 300c.how man

y calories does it lose to water? (specific heat capacity of aluminum is .21 cal/g0c)

1 answer:

Dmitry [639]3 years ago

6 0

Q = mCΔT
Q is heat, m is mass, C is specific heat, and ΔT is change in temp

Q = (50g)(.21)(500 degrees) = 5250 calories

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Which of the following would not be considered an observation in terms of the scientific method?

prohojiy [21]

Answer:

the first option, tasting a pasta sauce after adding a new ingredient.

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tasting a pasta sauce after adding a new ingredient is not an observation because there is no qualitative or quantitative data to be taken from that experience.

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Explanation:

Mole ratio of Oxygen to Hydrogen gas = 1 : 2.

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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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