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2 years ago

What is the mass in grams of 5.90 mol C8H18?

Chemistry

1 answer:

Radda [10]2 years ago

6 0

Answer:

Mass = 5.90 x 114 = 672.6g.

This can be rounded to 673g

C = 12 grams x 8 = 96 grams

H = 1 gram x 18 = 18 grams

Total = 96 + 18 = 114 grams in one mole.

If you have 5.9 moles, multiply this by the weight of one mole (114 grams) = 672.6 rounded to 673 grams in 5.9 moles.

Explanation:

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What are the moles of gas are in a 30 liter scuba canister if the temperature of the canister is 300 K and the pressure is 200 a

Lady_Fox [76]

Answer:

$n = 243.605$

Explanation:

Given

$Pressure, P = 200\ atm$

$Temperature, T = 300k$

$Volume, V = 30L$

Required

Calculate the number of moles

We'll apply the following formula to solve this question

$n = \frac{pV}{RT}$

Where

$R = 0.0821 L\ atm/(mol\ K)$

The above equation is an illustration of the ideal gas law

Substitute values for p, V, R and T in:

$n = \frac{pV}{RT}$

$n = \frac{200 * 30}{0.0821 * 300}$

$n = \frac{6000}{24.63}$

$n = 243.605$

Hence, there are 243.605 moles

7 0

3 years ago

One mole of carbon will have the same number of particles as one mole of hydrogen.

AVprozaik [17]

And the answer is False,it does not have the same number with particles as the one mole of hydrogen.

3 0

3 years ago

Read 2 more answers

When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2

MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is $\Delta \ T_f = T_{solvent}- T_{solution}$

$\Delta \ T_f = 2.9 ^0 \ C$

Using the depression in freezing point, the molar depression constant of the solvent $K_f = \dfrac{\Delta T_f}{m}$

$K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}$

$K_f = 4.82 ^0 C / m}$

The freezing point of the solution $\Delta T_f = T_{solvent} - T_{solution}$

$\Delta T_f = 7.3^ 0 \ C$

The molality of the solution is:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

Molar depression constant of solvent X, $K_f = 4.82 ^0 \ C/m$

Hence, using the elevation in boiling point;

the Vant'Hoff factor $i = \dfrac{\Delta T_f}{k_f \times m}$

$i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}$

$\mathbf {i = 1.51 }$

3 0

3 years ago

Calculate the molality of a solution that contains 51.2 g of naphthalene, C10H8, in 500 mL of carbon tetrachloride. The density

PtichkaEL [24]

Answer: The molality of naphthalene solution is 0.499 m

Explanation:

Density is defined as the ratio of mass and volume of a substance.

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$ ......(1)

Given values:

Volume of carbon tetrachloride = 500 mL

Density of carbon tetrachloride = 1.60 g/mL

Putting values in equation 1, we get:

$\text{Mass of carbon tetrachloride}=(1.60g/mL\times 500mL)=800g$

Molality is defined as the amount of solute expressed in the number of moles present per kilogram of solvent. The units of molarity are mol/kg. The formula used to calculate molarity:

$\text{Molality of solution}=\frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent (in g)}}$ .....(2)

Given values:

Given mass of naphthalene = 51.2 g

Molar mass of naphthalene = 128.17 g/mol

Mass of solvent = 800 g

Putting values in equation 2, we get:

$\text{Molality of naphthalene}=\frac{51.2\times 1000}{128.17\times 800}\\\\\text{Molality of naphthalene}=0.499m$

Hence, the molality of naphthalene solution is 0.499 m

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3 years ago

The brain tells his stomach intestines and liver want to be active and want to rest. Messages to and from the brain decide when

Nana76 [90]

Answer:

Nervous System I believe

Explanation:

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3 years ago