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3 years ago

How many cm in 725 km?

1 answer:

5 0

1 km is equal to 100,000 cm
$725km \times 100000cm \\ = 72500000 \: km .cm$
and you need to divide it to 1 km so that you can get rid of the km.
$= 72500000cm$

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What is the molar mass of HCl??

irakobra [83]

HCl = 1 + 35.5 => 36.5 g/mol

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4 years ago

Which of the following is an example of a consumer?

PIT_PIT [208]

Answer:

D. Rabbit

Explanation:

The rabbit is the consumer because he/she (im not judging) will have to consume other plants or small insects to get his/her (again not judging) energy

5 0

3 years ago

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How many molecules of ethane are present in 64.28 liters of ethane gas (C2H6) at STP

Rom4ik [11]

Answer : The molecule of ethane present in 64.28 L of ethane gas at STP is, $17.277\times 10^{23}$ molecule.

Solution :

At STP,

22.4 L volume of ethane present in 1 mole of ethane gas

64.28 L volume of ethane present in $\frac{64.28L}{22.4L}\times 1mole=2.869moles$ of ethane gas

And, as we know that

1 mole of ethane molecule contains $6.022\times 10^{23}$ molecules of ethane

2.869 moles of ethane molecule contains $2.869\times 6.022\times 10^{23}=17.277\times 10^{23}$ molecules of ethane

Therefore, the molecule of ethane present in 64.28 L of ethane gas at STP is, $17.277\times 10^{23}$ molecule.

4 0

3 years ago

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Which questions would help determine whether an article is scientifically accurate?

-BARSIC- [3]

I think it would be:
Is the author stating fact or opinion?
Does the author have evidence for the given facts?

Maybe
How many article has the author written before?

Ones you add if you wanted to is
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3 0

3 years ago

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Be sure to answer all parts. one of the most important industrial sources of ethanol is the reaction of steam with ethene derive

lions [1.4K]

Answer: 2.17x10⁻³ atm

Explanation:

First, we must write the balanced chemical equation for the process:

C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g)

The chemical reactions that occur in a closed container can reach a state of chemical equilibrium that is characterized because the concentrations of the reactants and products remain constant over time. The equilibrium constant of a chemical reaction is the value of its reaction quotient in chemical equilibrium.

The equilibrium constant (K) is expressed as the ratio between the molar concentrations (mol/L) of reactants and products. Its value in a chemical reaction depends on the temperature, so it must always be specified.

We will use the the equilibrium constant Kc of the reaction to calculate partial pressure of ethene. The constant Kc for the above reaction is,

Kc = $\frac{[C_{2} H_{5}OH]}{[H_{2}O][C_{2} H_{4}]}$

According to the law of ideal gases,

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the gas constant (0.082057 atm L / mol K) .

We can use the ideal gas law to determine the molar concentrations ([x] = n / V) from the gas pressures of ethanol and water, assuming that all gases involved behave as ideal gases. In this way,

PV = nRT → P = (n/V) RT → P = [x] RT → [x] = P / RT

So,

$[C_{2} H_{5}OH] = \frac{200 atm}{0.082057 \frac{atm L}{mol K} x 600 K } = 4.06 \frac{mol}{L}$

$[H_{2}O] = \frac{400 atm}{0.082057 \frac{atm L}{mol K} x 600 K } = 8.12 \frac{mol}{L}$

So, the molar concentration of ethene (C₂H₄) will be,

$[C_{2} H_{4}] = \frac{[C_{2} H_{5}OH]}{[H_{2}O] x Kc} = \frac{4.06 \frac{mol}{L} }{8.12 \frac{mol}{L}x9.00 x 10^{3} \frac{L}{mol} } = 5.56 x 10^{-5}\frac{mol}{L}$

Then, according to the law of ideal gases,

$P_{C_{2} H_{4}} = [C_{2} H_{4}]RT = 5.56 x 10^{-5} \frac{mol}{L} x 0.082057 \frac{atm L}{mol K} x 600 K = 2.17x10^{-3} atm$

So, when the partial pressure of ethanol is 200 atm and the partial pressure of water is 400 atm, the partial pressure of ethene at 600 K is 2.17x10⁻³ atm.

7 0

3 years ago