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mote1985 [20]
3 years ago
7

Please make the following calculation and round your

Chemistry
1 answer:
lapo4ka [179]3 years ago
3 0

Answer:

343

Explanation:

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3 years ago
How many grams of H2O can be made from the combustion of 3.75 liters of C7H14 and an excess of O2 at STP?
Kitty [74]

Answer:

21.10g of H2O

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2C7H14 + 21O2 —> 14CO2 + 14H2O

From the balanced equation above, 2L of C7H14 produced 14L of H2O.

Therefore, 3.75L of C7H14 will produce = (3.75 x 14)/2 = 26.25L of H2O.

Next, we shall determine the number of mole of H2O that will occupy 26.25L at stp. This is illustrated below:

1 mole of a gas occupy 22.4L at stp

Therefore, Xmol of H2O will occupy

26.25L i.e

Xmol of H2O = 26.25/22.4

Xmol of H2O = 1.172 mole

Therefore, 1.172 mole of H2O is produced from the reaction.

Next, we shall convert 1.172 mole of H2O to grams. This is illustrated below:

Number of mole H2O = 1.172 mole

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O =..?

Mass = mole x molar mass

Mass of H2O = 1.172 x 18

Mass of H2O = 21.10g

Therefore, 21.10g of H2O is produced from the reaction.

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3 years ago
At a certain temperature, the solubility of n2 gas in water at 3.08 atm is 72.5 mg of n2 gas/100 g water . calculate the solubil
8090 [49]

According to Henry's law, solubility of solution is directly proportional to partial pressure thus,

\frac{S_{1}}{P_{1}}=\frac{S_{2}}{P_{2}}

Solubility at pressure 3.08 atm is 72.5/100, solubility at pressure 8 atm should be calculated.

Putting the values in equation:

\frac{0.725}{3.08}=\frac{S_{2}}{8}

On rearranging,

S_{2}=\frac{0.725\times 8}{3.08}=1.88

Therefore, solubility will be 1.88 mg of N_{2} gas in 1 g of water or, 188 mg of tex]N_{2}[/tex] gas in 100 g of water.

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cestrela7 [59]
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3 years ago
Read 2 more answers
The gas law for an ideal gas at absolute temperature T (in kelvins), pressure P (in atmospheres), and volume V (in liters) is PV
allsm [11]

Answer:

The rate of change of the temperature is 0.0365 Kelvin per minute.

Explanation:

<u>Step 1</u>: Given data

ideal gas law: P*V = n*R*T

with P= pressure of the gas ( in atm) = 9.0 atm

with V= volume of the gass (in L) =12L

with n = number of moles = 10 moles

R = gas constant = 0.0821 L*atm* K^−1*mo^−1

T = temperature = TO BE DETERMINED

The volume  decreases with a rate of 0.17L/min = dV/dT  = -0.17

The pressure increases at a rate of 0.13atm/min = dP/dT

<u>Step 2:</u> The ideal gas law

P * [dV/dT] + V * [dP/dT] = nR * dT/dt

9 atm * (-0.17L/min) + 12L * 0.13atm/min = 10 moles * 0.0821 L*atm* K^−1*mo^−1 *dT/dt

0.03 = 0.821 * dT/dt

dT/dt = 0.03/0.821

dT/dt = 0.0365

Since the gas constant is expressed in Kelvin and not in °C, this means that <u>the rate of chagnge of the temperature is 0.0365 Kelvin per 1 minute.</u>

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3 years ago
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