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Flura [38]
3 years ago
12

If you have a gas sample with a pressure of 2 atm, a volume of 40 L and a temperature of 300 K, what will be the number of moles

of gas in that sample?
Chemistry
1 answer:
Karolina [17]3 years ago
6 0

Answer:

n = 3.25 mol

Explanation:

Given data:

Pressure of sample = 2 atm

Volume of sample = 40 L

Temperature = 300 K

Number of moles = ?

Solution:

PV = nRT

R = general gas constant = 0.0821 atm.L/mol.K

2 atm ×40 L = n × 0.0821 atm.L/mol.K × 300 K

80 atm.L = n  × 24.63 atm.L/mol

n = 80 atm.L  /24.63 atm.L/mol

n = 3.25 mol

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What is the percent composition of carbon in glucose (C6H12O6)? (The molar mass of C = 12.01, H = 1.0079 and O = 16.00.)
igomit [66]
The molar mass of C6H12O6 is 180.1548 g/mol. The molar mass of C6 is 72.06 g/mol. So you divide the molar mass of C6 by the molar mass of C6H12O6 and multiply it by 100: (72.06g C6/180.1548g C6H1206)x100= .3999 x 100= 39.99% Carbon in glucose (40% rounded) 
6 0
3 years ago
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ella [17]

Answer:

i honestly dont know

Explanation:

8

7 0
3 years ago
determine the empirical and molecular formula of a compound composed of 18.24 g carbon, 0.51 g hydrogen, and 16.91 g fluorine an
Vika [28.1K]

Answer: The empirical formula for the given compound is C_3HF_2  and molecular formula for the given compound is C_{24}H_8F_{16}

Explanation : Given,

Mass of C = 18.24 g

Mass of H = 0.51 g

Mass of F = 16.91 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{18.24g}{12g/mole}=1.52moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.51g}{1g/mole}=0.51moles

Moles of Fluorine = \frac{\text{Given mass of Fluorine}}{\text{Molar mass of Fluorine}}=\frac{16.91g}{19g/mole}=0.89moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.51 moles.

For Carbon = \frac{1.52}{0.51}=2.98\approx 3

For Hydrogen  = \frac{0.51}{0.51}=1

For Fluorine = \frac{0.89}{0.51}=1.74\approx 2

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : F = 3 : 1 : 2

The empirical formula for the given compound is C_3H_1F_2=C_3HF_2

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{molecular mass}}{\text{empirical mass}}

We are given:

Molar mass  = 562.0 g/mol

Mass of empirical formula = 3(12) + 1(1) + 2(19) = 75 g/eq

Putting values in above equation, we get:

n=\frac{562.0}{75}=7.49\approx 8

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_3HF_2=(C_3HF_2)_n=(C_3HF_2)_8=C_{24}H_8F_{16}

Thus, the molecular formula for the given compound is C_{24}H_8F_{16}

8 0
3 years ago
A solution contains 0.115 mol h2o and an unknown number of moles of nacl. the vapor pressure of the solution at 30°c is 25.7 tor
Valentin [98]
According to Raoult's low:
We will use this formula: Vp(Solution) = mole fraction of solvent * Vp(solvent)
∴ mole fraction of solvent = Vp(Solu) / Vp (Solv)
when we have Vp(solu) = 25.7 torr & Vp(solv) = 31.8 torr 
So by substitution:
∴ mole fraction of solvent = 25.7 / 31.8 =0.808 
when we assume the moles of solute NaCl = X 
and according to the mole fraction of solvent formula:
mole fraction of solvent = moles of solvent / (moles of solvent + moles of solute)
by substitute:
∴ 0.808 = 0.115 / (0.115 + X)
 So X (the no.of moles of NaCl) = 0.027 m
8 0
3 years ago
When solution A was mixed with solution B, the solution turned cloudy, the test tube did not feel warm, and there were no visibl
son4ous [18]

Answer:

Precipitation reaction

Explanation:

Given that solution A was mixed with solution B, the solution turned cloudy. The test is not warm and no bubbles visible. This means that the precipiate is formed.

The concept is when two colourless solutions react to form a cloudy precipitate that settles at bottom of a solution then the reaction is said to be a precipitation reaction.

An example can be the Reaction of Silver nitrate with common salt.

3 0
3 years ago
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