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xz_007 [3.2K]
3 years ago
15

What is the density of CO2 at a pressure of 0.0079 atm and 227 K? (These are the approximate atmospheric in Mars)

Chemistry
1 answer:
Strike441 [17]3 years ago
6 0

Answer:

The density is 0.0187 g/L

Explanation:

First thing to do here is to calculate the Volume of 1 mole of CO2 using the ideal gas equation

Mathematically;

PV = nRT

thus V = nRT/P

what we have are;

n = 1 mole

R is the molar has constant = 0.082 L•atm•mol^-1•K^-1

P is the pressure = 0.0079 atm

T is temperature = 227 K

Substituting these values, we have;

V = nRT/P = (1 * 0.082 * 227)/0.0079

V = 2,356.20 dm^3

This means according to the parameters given in the question, the volume of 1 mole of carbon iv oxide is 2,356.20 dm^3

But this is not what we want to calculate

What we want to calculate is the density

Mathematically, we can calculate the density using the formula below;

density = molar mass/molar volume

Kindly recall that the molar mass of carbon iv oxide is 44 g/mol

Thus the density = 44/2356.20 = 0.018674136321195 which is approximately 0.0187 g/L

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Answer: 406 hours

Explanation:

Q=I\times t

where Q= quantity of electricity in coloumbs

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t= time in seconds = ?

The deposition of copper at cathode is represented by:

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96500\times 2=193000 Coloumb of electricity deposits 1 mole of copper

i.e. 63.5 g of copper is deposited by = 193000 Coloumb

Thus 19.0 kg or 19000 g of copper is deposited by = \frac{193000}{63.5}\times 19000=57748032 Coloumb

57748032=39.5\times t

t=1461975sec=406hours    (1hour=3600s)

Thus it will take 406 hours to plate 19.0 kg of copper onto the cathode if the current passed through the cell is held constant at 39.5 A

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WHY IS THE OCEAN SALTY ALSO I CANT TAKE OF CAP LOCKS ON A CROME BOOK
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Explanation:

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Un anillo, de masa 90 gramos, contiene 59,1% de oro. ¿Cuál es el valor del anillo si cada mol de oro vale S/.1800?. Considere el
LekaFEV [45]

Answer:

S/.486 es el valor del anillo

Explanation:

Para hallar el precio del anillo se deben encontrar las moles de oro que contiene este.

Si el anillo es de 90g y solo el 59.1% contiene oro, la cantidad de oro en gramos es:

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Ahora, para convertir los gramos de oro a moles se debe usar la masa atómica del oro (197g/mol), así:

53.19g × (1mol / 197g) = <em><u>0.27 moles de oro contiene el anillo</u></em>.

Ya que cada mol de oro cuesta S/.1800, 0.27 moles de oro (Y por lo tanto, el anillo) costarán:

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This part insulates the reaction chamber from the transfer of heat to or from the surrounding environment. this part contains a
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Therefore, part A measures the temperature of the reaction mixture.

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