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4 years ago

determine the final temperature of a 10-0-g aluminum block originally at 25.0°C if you apply 435 joules of energy to it

1 answer:

mihalych1998 [28]4 years ago

5 0

Energy = Mass * heat capacity * temperature change so,
The energy added is 435 J and the temperature has to increase since the energy is added.

435 J = 10.0 g * 0.89 J/gC * temperature change 

Temperature change = 48.9 C 

The initial temperature is 25.0 C, the final temperature is 25.0 C + 48.9 C = 73.9 C.

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Which weather technology can detect rain in tropical storms and hurricanes that are within about 250 miles, estimate how much ra

masha68 [24]

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3 years ago

The volume of a gas-filled balloon is 30.0 L at 40 °C and 153 kPa pressure. What volume will the balloon have at standard temper

soldier1979 [14.2K]

Hello!

The volume of a gas-filled balloon is 30.0 L at 40 °C and 153 kPa pressure. What volume will the balloon have at standard temperature and pressure (273.15 K and 101.3 kPa)?

a. 17.3 L

b. 23.7 L

c. 39.5 L

d. 51.9 L

We have the following data:

V1 (initial volume) = 30 L

T1 (initial temperature) = 40ºC (in Kelvin)

TK = TºC + 273.15

TK = 40 + 273.15 → T1 (initial temperature) = 313.15 K

P1 (initial pressure) = 153 kPa

V2 (final volume) = ? (in L)

T2 (final temperature) = 273.15 K

P2 (final pressure) = 101.3 kPa

Now, we apply the data of the variables above to the General Equation of Gases, let's see:

$\dfrac{P_1*V_1}{T_1} =\dfrac{P_2*V_2}{T_2}$

$\dfrac{153*30}{313.15} =\dfrac{101.3*V_2}{273.15}$

$\dfrac{4590}{313.15} =\dfrac{101.3\:V_2}{273.15}$

multiply the means by the extremes

$313.15*101.3\:V_2 = 4590*273.15$

$31722.095\:V_2 = 1253758.5$

$V_2 = \dfrac{1253758.5}{31722.095}$

$\boxed{\boxed{V_2 \approx 39.5\:L}}\:\:\:\:\:\:\bf\blue{\checkmark}$

Answer:

c. 39.5 L

_______________________

A gas that has a volume of 28 liters, a temperature of 45 °C, and an unknown pressure, has its volume increased to 34 liters and its temperature decreased to 35 °C. If I measure the pressure after the change to be 2.0 atm, what was the original pressure of the gas?

a. 1.5 atm

b. 1.7 atm

c. 2.8 atm

d. 2.5 atm

We have the following data:

V1 (initial volume) = 28 L

T1 (initial temperature) = 45ºC (in Kelvin)

TK = TºC + 273.15

TK = 45 + 273.15 → T1 (initial temperature) = 318.15 K

P1 (initial pressure) = ? (in atm)

V2 (final volume) = 34 L

T2 (final temperature) = 35ºC (in Kelvin)

TK = TºC + 273.15

TK = 35 + 273.15 → T2 (final temperature) = 308.15 K

P2 (final pressure) = 2 atm

Now, we apply the data of the variables above to the General Equation of Gases, let's see:

$\dfrac{P_1*V_1}{T_1} =\dfrac{P_2*V_2}{T_2}$

$\dfrac{P_1*28}{318.15} =\dfrac{2*34}{308.15}$

$\dfrac{28\:P_1}{318.15} =\dfrac{68}{308.15}$

multiply the means by the extremes

$28\:P_1*308.15 = 318.15*68$

$8628.2\:P_2 = 21634.2$

$P_2 = \dfrac{21634.2}{8628.2}$

$P_2 = 2.507382768... \to \boxed{\boxed{P_2 \approx 2.5\:atm}}\:\:\:\:\:\:\bf\blue{\checkmark}$

Answer:

d. 2.5 atm

_______________________

$\bf\green{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}$

3 0

3 years ago

In an experiment, a 0.5297 g sample of diphenylacetylene (C14H10) is burned completely in a bomb calorimeter. The calorimeter is

V125BC [204]

Answer:

the Molar heat of Combustion of diphenylacetylene $(C_{14}H_{10})$ = $-6.931 *10^3 \ kJ/mol$

Explanation:

Given that:

mass of diphenylacetylene $(C_{14}H_{10})$ = 0.5297 g

Molar Mass of diphenylacetylene $(C_{14}H_{10})$ = 178.21 g/mol

Then number of moles of diphenylacetylene $(C_{14}H_{10})$ = $\frac{mass}{molar \ mass}$

= $\frac{0.5297 \ g }{178.24 \ g/mol}$

= 0.002972 mol

By applying the law of calorimeter;

Heat liberated by 0.002972 mole of diphenylacetylene $(C_{14}H_{10})$ = Heat absorbed by $H_2O$ + Heat absorbed by the calorimeter

Heat liberated by 0.002972 mole of diphenylacetylene $(C_{14}H_{10})$ = msΔT + cΔT

= 1369 g × 4.184 J g⁻¹°C⁻¹ × (26.05 - 22.95)°C + 916.9 J/°C (26.05 - 22.95)°C

= 17756.48 J + 2842.39 J

= 20598.87 J

Heat liberated by 0.002972 mole of diphenylacetylene $(C_{14}H_{10})$ = 20598.87 J

Heat liberated by 1 mole of diphenylacetylene $(C_{14}H_{10})$ will be = $\frac{20598.87 \ J}{0.002972 \ mol}$

= 6930979.139 J/mol

= 6930.98 kJ/mol

Since heat is liberated ; Then, the Molar heat of Combustion of diphenylacetylene $(C_{14}H_{10})$ = $-6.931 *10^3 \ kJ/mol$

3 0

3 years ago

Read 2 more answers

How do you calculate the number of protons, neutrons, and electrons in an element?

Nezavi [6.7K]

Answer:

The first thing you will need to do is find some information about your element. Go to the Periodic Table of Elements and click on your element. If it makes things easier, you can select your element from an alphabetical listing.

Number of Protons = Atomic Number

Number of Electrons = Number of Protons = Atomic Number

Number of Neutrons = Mass Number - Atomic Number

For krypton:

Number of Protons = Atomic Number = 36

Number of Electrons = Number of Protons = Atomic Number = 36

Number of Neutrons = Mass Number - Atomic Number = 84 - 36 = 48

Explanation:

hope this helps, have a good day :-)

3 0

3 years ago

Which power source uses the earth’s internal heat?

Nataly_w [17]

Answer:

Its geothermal energy.

Explanation:

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5 0

4 years ago