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Kisachek [45]
3 years ago
14

If y > z, which of the following equations must be true?

Chemistry
1 answer:
marin [14]3 years ago
8 0

Answer:

I think the answer is y-z>0

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I don't know how to do any of these.plz help.its a study guide for my chemistry exam tomorrow ✌
Vladimir [108]

6.02 x 1023 atoms weigh out 63.55 grams copper.


No. of Molecules in water = 3.5mole x (6.02 x 10^23) molecules/mole = 2.107 x 10^24 molecules of H2O

8 0
3 years ago
When a 0.245-g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises 1.643 ∘C . When a 0.260-g sample
sveticcg [70]

Answer:

The heat of combustion per moles of caffeine is 4220 kJ/mol

Explanation:

Step 1: Data given

⇒ When  benzoic acid sample of 0.245 grams is burned the temperature rise is 1.643 °C

⇒ When 0.260 gram of caffeine is burned, the temperature rise is 1.436 °C

⇒ Heat of combustion of benzoic acid = 26.38 kJ/g

<u>Step 2:</u> Calculate the heat released: for combustion of benzoic acid

0.245 g benzoic acid *  26.38 kJ/g = 6.4631 kJ

<u>Step 3</u>: Calculate the heat capacity of the calorimeter:

c = Q/ΔT

Q = 6.4631 kJ   / 1.643°C = 3.934 kJ/ °C

<u>Step 4:</u> Calculate moles of a 0.260 g sample of caffeine:

Moles caffeine = Mass caffeine / Molar mass caffeine

0.260 grams/ 194.19 g/mol  = 0.0013389 moles

Step 5: Calculate heat released: for combustion of caffeine

Q = c * ΔT

Q = 3.934 kJ/°C * 1.436 °C = 5.65 kJ

Step 6: Calculate the heat of combustion per mole of caffeine  

5.65 kJ  /  0.0013389 moles = 4219.9 kJ/mol  ≈ 4220 kJ/mol

The heat of combustion per moles of caffeine is 4220 kJ/mol

4 0
3 years ago
pplication)Using multiple models simultaneously: This FNT refers to a processinvolving three moles of a diatomic gas (which beha
Fiesta28 [93]

Complete Question

Questions Diagram is attached below

Answer:

*  W=1142.86Joule

*  Q=997.7J

*  H=2140.5J

Explanation:

From the question we are told that:

Temperature T=337K

Pressure P=(60-55)Pa*10^5

VolumeV=(1.6-1.4)m^3*10^{-3}

Generally the equation for gas Constant is mathematically given by

\frac{P_2}{P_1}=\frac{V_1}{V_2}^n

 \frac{55*10^5}{60*10^5}=\frac{1.4*10^{-3}}{1.6*10^{-3}}^n

 n=0.65

Therefore

Work-done

 W=\int{pdv}

 W=\frac{55*10^5*1.6*10^{-3}*60*10^5*1.4*10^{-3}}{1-0.65}

 W=1142.86Joule

Generally the equation for internal energy is mathematically given by

 Q=mC_vdT\\\\Q=\frac{3*1*3.314*16}{1.4-1}

 Q=997.7J

Therefore

 H=Q+W

 H=997.7J-11.42.9

 H=2140.5J

5 0
2 years ago
The vapor pressure of water at 65oC is 187.54 mmHg. What is the vapor pressure of a ethylene glycol (CH2(OH)CH2(OH)) solution ma
Pavlova-9 [17]

Answer:

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

Explanation:

Vapor pressure of water at 65 °C=p_o= 187.54 mmHg

Vapor pressure of the solution at 65 °C= p_s

The relative lowering of vapor pressure of solution in which non volatile solute is dissolved is equal to mole fraction of solute in the solution.

Mass of ethylene glycol = 22.37 g

Mass of water in a solution = 82.21 g

Moles of water=n_1=\frac{82.21 g}{18 g/mol}=4.5672 mol

Moles of ethylene glycol=n_2=\frac{22.37 g}{62.07 g/mol}=0.3603 mol

\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}

\frac{187.54 mmHg-p_s}{187.54 mmHg}=\frac{0.3603 mol}{0.3603 mol+4.5672 mol}

p_s=173.83 mmHg

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

6 0
2 years ago
What happens when nitrogen and hydrogen comes in contact at high temperature and pressure in presence of iron powder and molybde
Sidana [21]

Answer:

<h2>Ammonia Gas</h2>

Explanation:

It result in formation of ammonia gas.

N2 + 3H2 ---<u>iron</u><u>/</u><u>molybdenum</u><u>/</u><u>high</u><u> </u>temp/pres--- > 2 NH3

It forms ammonia gas.

Please mark branliest if you are satisfied with the answer. Thanking in anticipation.

6 0
3 years ago
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