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3 years ago

Help Anyone! What would be the correct answer to this question?

Chemistry

1 answer:

Ymorist [56]3 years ago

8 0

I have no clue..........

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After an afternoon party, a small cooler full of ice is dumped onto the hot ground and melts. If the cooler contained 6.60 kg of

gtnhenbr [62]

Answer:

The quantity of heat required to melt all the ice at 0°C is 2.21 * 10⁶ J

Explanation:

Latent heat of fusion is the heat absorbed by a unit mass of a given solid at its melting point that completely converts the solid to a liquid at the same temperature. Its unit is Joules/kg or Joules/g.

1 calorie = 4.184 Joules

Therefore , 80.0 cal/g = 80.0 cal/g * 4.184 J/cal = 334.72 J/g

1 g = 0.001 kg; Heat of fusion in J/kg = 334.72 J/g * 1g /0.001 kg = 3.35 * 10⁵ J/kg

Quantity of heat, Q = mass * latent heat of fusion of ice

quantity of heat required = 6.60 kg * 3.35 * 10⁵ J/kg

Quantity of heat required = 2.21 * 10⁶ J

Therefore, the quantity of heat required to melt all the ice at 0°C is 2.21 * 10⁶ J

5 0

3 years ago

Calculate the mass of methane that must be burned to provide enough heat to convert 242.0 g of water at 26.0°C into steam at 101

Anarel [89]

Answer: The mass of methane burned is 12.4 grams.

Explanation:

The chemical equation for the combustion of methane follows:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CO_2(g))})+(2\times \Delta H^o_f_{(H_2O(g))})]-[(1\times \Delta H^o_f_{(CH_4(g))})+(2\times \Delta H^o_f_{(O_2(g))})]$

We are given:

$\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.51kJ/mol\\\Delta H^o_f_{(CH_4(g))}=-74.81kJ/mol\\\Delta H^o_f_{O_2}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-393.51))+(2\times (-241.82))]-[(1\times (-74.81))+(2\times (0))]\\\\\Delta H^o_{rxn}=-802.34kJ$

The heat calculated above is the heat released for 1 mole of methane.

The process involved in this problem are:

$(1):H_2O(l)(26^oC)\rightarrow H_2O(l)(100^oC)\\\\(2):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(3):H_2O(g)(100^oC)\rightarrow H_2O(g)(101^oC)$

Now, we calculate the amount of heat released or absorbed in all the processes.

For process 1:

$q_1=mC_p,l\times (T_2-T_1)$

where,

$q_1$ = amount of heat absorbed = ?

m = mass of water = 242.0 g

$C_{p,l}$ = specific heat of water = 4.18 J/g°C

$T_2$ = final temperature = $100^oC$

$T_1$ = initial temperature = $26^oC$

Putting all the values in above equation, we get:

$q_1=242.0g\times 4.18J/g^oC\times (100-(26))^oC=74855.44J$

For process 2:

$q_2=m\times L_v$

where,

$q_2$ = amount of heat absorbed = ?

m = mass of water or steam = 242 g

$L_v$ = latent heat of vaporization = 2257 J/g

Putting all the values in above equation, we get:

$q_2=242g\times 2257J/g=546194J$

For process 3:

$q_3=mC_p,g\times (T_2-T_1)$

where,

$q_3$ = amount of heat absorbed = ?

m = mass of steam = 242.0 g

$C_{p,g}$ = specific heat of steam = 2.08 J/g°C

$T_2$ = final temperature = $101^oC$

$T_1$ = initial temperature = $100^oC$

Putting all the values in above equation, we get:

$q_3=242.0g\times 2.08J/g^oC\times (101-(100))^oC=503.36J$

Total heat required = $q_1+q_2+q_3=(74855.44+546194+503.36)=621552.8J=621.552kJ$

To calculate the number of moles of methane, we apply unitary method:

When 802.34 kJ of heat is needed, the amount of methane combusted is 1 mole

So, when 621.552 kJ of heat is needed, the amount of methane combusted will be = $\frac{1}{802.34}\times 621.552=0.775mol$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of methane = 16 g/mol

Moles of methane = 0.775 moles

Putting values in above equation, we get:

$0.775mol=\frac{\text{Mass of methane}}{16g/mol}\\\\\text{Mass of methane}=(0.775mol\times 16g/mol)=12.4g$

Hence, the mass of methane burned is 12.4 grams.

8 0

3 years ago

Which of the following best describes the change in Antarctic temperature from about 440,000 years ago to about 340,000 years ag

Vlada [557]

Answer: С . The temperature increases by about 12°C and then decreases by about 12°C.

Explanation:

Temperatures around the world have been on the rise since the Industrial revolution as humans clog the planet with Carbon Dioxide and other pollutants. This had led to a rise in temperatures that has seen ice levels fall and sea levels rise around the world.

Temperature fluctuations on the other hand are not a new thing. Studies show that in Antarctica temperatures from about 440,000 years ago to about 340,000 years ago increased by 12°C and then decreased by about 12°C.

7 0

3 years ago

Read 2 more answers

The energy in eV for light with a wavelength of 6250 angstroms is _. Note - there are 1.6 x 10-12 erg in 1 eV.

vivado [14]

Answer:

2 eV

Explanation:

The energy of a photon of light is given by the formula

$E=\frac{hc}{\lambda}$