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

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

Step 2: Calculate the heat released: for combustion of benzoic acid

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

Step 3: Calculate the heat capacity of the calorimeter:

c = Q/ΔT

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

Step 4: 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 $P=(60-55)Pa*10^5$

Volume $V=(1.6-1.4)m^3*10^{-3}$

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 ---iron/molybdenum/high 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