Do PV cells last for a long time and are efficient

A brand of dish soap says that it will get your dishes cleaner. What is the

vodka [1.7K]

Answer:

B

Explanation:

Si no demuestra que funciona te estarían estafando

3 0

3 years ago

The Haber Process synthesizes ammonia at elevated temperatures and pressures. Suppose you combine 1580 L of nitrogen gas and 351

ikadub [295]

Answer : The volume of reactant measured at STP left over is 409.9 L

Explanation :

First we have to calculate the moles of $N_2$ and $H_2$ by using ideal gas equation.

$PV_{N_2}=n_{N_2}RT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of $N_2$ gas = 1580 L

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times 1580L=n_{N_2}\times (0.0821L.atm/mol.K)\times 273K$

$n_{N_2}=70.49mole$

$PV_{H_2}=n_{H_2}RT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of $H_2$ gas = 3510 L

n = number of moles $H_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times 3510L=n_{H_2}\times (0.0821L.atm/mol.K)\times 273K$

$n_{H_2}=156.6mole$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

From the balanced reaction we conclude that

As, 3 mole of $H_2$ react with 1 mole of $N_2$

So, 156.6 moles of $H_2$ react with $\frac{156.6}{3}\times 1=52.2$ moles of $N_2$

From this we conclude that, $N_2$ is an excess reagent because the given moles are greater than the required moles and $H_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the excess moles of $N_2$ reactant (unreacted gas).

Excess moles of $N_2$ reactant = 70.49 - 52.2 = 18.29 moles

Now we have to calculate the volume of reactant, measured at STP, is left over.

$PV=nRT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of gas = ?

n = number of moles of unreacted gas = 18.29 moles

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times V=18.29mole\times (0.0821L.atm/mol.K)\times 273K$

$V=409.9L$

Therefore, the volume of reactant measured at STP left over is 409.9 L

8 0

3 years ago

Why does hot chocolate powder dissolve faster in hot water than cold water

Bezzdna [24]

Because, particles are moving faster in the hotter material. This speeds up the dissolving process.

6 0

3 years ago

g . Calculate the molar concentration for each of the following solutions. (a) 1.50 g NaCl in 100.0 mL of solution (b) 1.50 g K2

oksano4ka [1.4K]

Answer:

(a) $M=0.257M$

(b) $M=0.0510M$

(c) $M =0.500M$

(d) $M= 0.391M$

Explanation:

Hello,

In this case, since the molarity or molar concentration of a solution is computed by dividing the moles of solute by the volume of solution in liters, we proceed as follows:

(a) The molar mass of sodium chloride is 58.45 g/mol and the volume in liters is 0.100 L, therefore, the molarity is:

$M=\frac{1.50gNaCl}{0.100L} *\frac{1molNaCl}{58.45gNaCl} =0.257M$

(b) The molar of potassium dichromate is 294.2 g/mol and the volume in liters is 0.100 L, therefore, the molarity is:

$M=\frac{1.50gK_2Cr_2O_7}{0.100L} *\frac{1molK_2Cr_2O_7}{294.2gK_2Cr_2O_7} =0.0510M$

(c) The molar of calcium chloride is 111 g/mol and the volume in liters is 0.125 L, therefore, the molarity is:

$M=\frac{5.55gCaCl_2}{0.100L} *\frac{1molCaCl_2}{111gCaCl_2} =0.500M$

(d) The molar of sodium sulfate is 142 g/mol and the volume in liters is 0.125 L, therefore, the molarity is:

$M=\frac{5.55gNa_2SO_4}{0.100L} *\frac{1molNa_2SO_4}{142gNa_2SO_4} = 0.391M$

Best regards.

4 0

3 years ago

How well can you apply Charles’s law to this sample of gas that experiences changes in pressure and volume? Assume that pressure

GarryVolchara [31]

Answer:

A: 384

B: 0.85

C: 1.1

D: 221

Explanation:

8 0

4 years ago

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