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navik [9.2K]
3 years ago
9

In examples where the volume of a system is under constant pressure, which of

Chemistry
1 answer:
grin007 [14]3 years ago
6 0
Answer:

D


Explanation:


At constant volume, the heat of reaction is equal to the change in the internal energy of the system. ... Most chemical reactions occur at constant pressure, so enthalpy is more often used to measure heats of reaction than internal energy.
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How does more greenhouse gases in the Earth's atmosphere affect earth's temperature?
Mumz [18]

Explanation:

Greenhouse gasses in Earth's atmosphere affect Earth's temperature by making it hotter. Greenhouse gasses make the Earth warmer by trapping heat. You can think of greenhouse gasses as Earth's blanket.

5 0
3 years ago
Read 2 more answers
A 125g sample of a mixture of salt and sand contains 12% salt.
natka813 [3]
12 % salt is present in 125 g mixture of salt and sand.
Keep in mind that the total percentage is always 100 %
Therefore, if 12 % is the salt, remaining 88 % must be
sand.
a. The amount of mixture is 125 q. Here, 12 % of 125 is
12 * 125 / 100 = 15 g of salt is present in 125 g mixture.
b. The amount of sand can be calculated similarly, 88 %
of 125 g is 88 * 125 / 100 = 110 g of sand is present in
125 g mixture.
4 0
3 years ago
Calculate Ho298 for the process
Inga [223]

Explanation:

As per the Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

Hence, according to this law the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

Sb + \frac{3}{2}Cl_2 \rightarrow SbCl_{3}    \Delta H^0_1 =  -314 kJ  ..........(1)

SbCl_{3} + Cl_2 \rightarrow SbCl_{5}    \Delta H^0_2 = -80kJ   ..............(2)

The final reaction is as follows:  

Sb + \frac{5}{2}Cl_{2} \rightarrow SbCl_{5}  \Delta H^0_3 = ?  .............(3)

Therefore, adding (1) and (2) we get the final equation (3) and value of \Delta H^{0}_{3} at 298 K will be as follows.

             \Delta H^{0}_{3} = \Delta H^{0}_{1} + \Delta H^{0}_{2}    

                       = -314 kJ + (-80) kJ

                       = -394 kJ

Thus, we can conclude that H^{o} at 298 K for the given process is -394 kJ.

4 0
3 years ago
9) After lab, all of Darrel’s friends looked at his data and laughed and laughed. They told him that he was 30.8% too low in the
zaharov [31]
If he was 30.8% too low, it means that he was at 69.2% of the boiling point needed. So 50o C is 69.2% of total.

In order to know what 100% is, you can divide the number by it's percentage and then multiply it by a hundred.

So: 50/30.8=1.623
1.623*100=162.3
So the correct boiling point of the liquid he was working with in the lab is 162.3 oC
3 0
4 years ago
Read 2 more answers
Question 9
Evgesh-ka [11]

Answer:

0.382 atm

Explanation:

In order to find the pressure, you need to know the moles of carbon dioxide (CO₂) gas. This can be found by multiplying the mass (g) by the molar mass (g/mol) of CO₂. It is important to arrange the conversion in a way that allows for the cancellation of units.

Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)

Molar Mass (CO₂): 44.007 g/mol

15 grams CO₂               1 mole
----------------------  x  ------------------------  =  0.341 moles CO₂
                                44.007 grams

To find the pressure, you need to use the Ideal Gas Law equation.

PV = nRT

In this equation,

-----> P = pressure (atm)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas Constant (0.08206 atm*L/mol*K)

-----> T = temperature (K)

After you convert Celsius to Kelvin, you can plug the given and calculated values into the equation and simplify to find the pressure.

P = ? atm                              R = 0.08206 atm*L/mol*K

V = 20 L                               T = 0 °C + 273.15 = 273.15 K

n = 0.341 moles

PV = nRT

P(20 L) = (0.341 moles)(0.08206 atm*L/mol*K)(273.15 K)

P(20 L) = 7.64016

P = 0.382 atm

7 0
2 years ago
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