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4 years ago

How many grams of o2 are required to produce 7.00 g of co2?

1 answer:

7 0

Hey there!:

Molar mass:

O2 = 31.99 g/mol
CO2 = 44 g/mol

By the stoichiometry of the reaction:

C + O2 = CO2

31.99 g O2 --------------- 44 g CO2
mass O2 ------------------ 7.00 g CO2

mass O2 = ( 7.00 * 31.99 ) / 44

mass O2 = 223.93 / 44

mass O2 = 5.089 g

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A solution of a sugar with the chemical formula (c12h22o11) is prepared by dissolving 8.45 g in 250.0 ml of water at 25

lapo4ka [179]

A. Concentration in terms of molarity

First, calculate for the molar mass of the given solution, C12H22O11.
m = 12(12) + 22(1) + 11(16) = 232
Calculate for the number of moles.

n = 8.45 g / 232 g/mol = 0.036422 moles

Divide the determined number of moles by volume of solution in L.
V(solution) = (250 mL)(1 L/1000 mL) = 0.25 L

The concentration in molarity is determined by dividing the number of moles by the volume of solution in liters.
M = 0.03622 moles / 0.25 L = 0.146 moles/L

b. Concentration in terms of molality
Calculate the mass of the solvent in kilogram given the volume and its density.
m(solvent) = (1.21 g/cm3)(250 mL)(1 kg/1000 g) = 0.3025 kg
Divide the calculated number of moles by the mass of the solvent in kilograms.

molality = 0.036422 moles/0.3025 kg = 0.12 molal

c. Concentration in terms of weight percent.
The total weight of the solution is equal to 302.5 g.

weight percent of solute = (8.45 g / 302.5 g)(100%) = 2.79%

d. Weight percent in ppm.
weight percent of solute x (10000 ppm/1%)
= (2.79%)(10000 ppm/1%) = 27933.88 ppm

6 0

3 years ago

Read 2 more answers

Calculate the mass of a liquid with a density of 4.7 g/mL and a volume of 15 mL. *

Tems11 [23]

Mass = 70.5 grams
D = m/v
D=4.7 g/mL
V=15mL
4.7=m/15
4.7=1/15m
M=70.5g

3 0

3 years ago

A 2.50-l flask contains a mixture of methane (ch4) and propane (c3h8) at a pressure of 1.45 atm and 20°c. when this gas mixture

Cerrena [4.2K]

Answer:- Mole fraction of methane in the original gas mixture is 0.854.

Solution:- From given volume, pressure and temperature, we could calculate the total moles of the gaseous mixture of methane and propane using ideal gas law as:

PV = nRT

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

V = 2.50 L

P = 1.45 atm

T = 20 + 273 = 293 K

Let's plug in the values in the equation:

$n=(\frac{1.45*2.50}{0.0821*293})$

n = 0.151

Let's say the solution has X moles of methane. Then moles of propane would be = (0.151 - X)

The combustion equations of methane and propane are:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

From methane balanced equation, there is 1:1 mol ratio between methane and carbon dioxide. So, X moles of methane would produce X moles of carbon dioxide.

From balanced equation of propane, there is 1:3 mol ratio between propane and carbon dioxide. So, (0.151 - X) moles of propane would give 3(0.151 - X) moles of carbon dioxide.

So, total moles of carbon dioxide that we would get from methane and propane combustion are:

X + 3(0.151 - X)

From given data, 8.60 g of carbon dioxide are formed by the combustion of gas mixture.

moles of Carbon dioxide = $8.60g(\frac{1mol}{44g})$

moles of carbon dioxide = 0.195 mol

Hence, 0.195 = X + 3(0.151 - X)

Let's solve this for X as:

0.195 = X + 0.453 - 3X

0.195 = 0.453 - 2X

2X = 0.453 - 0.195

2X = 0.258

$X=\frac{0.258}{2}$

X = 0.129

So, there are 0.129 moles of methane in the mixture.

moles of propane = 0.151 - 0.129 = 0.022

mole fraction of methane = $\frac{moles of Methane}{total moles}$

mole fraction of methane = $\frac{0.129}{0.151}$

mole fraction of methane = 0.854

Hence, the mole fraction of methane gas in the original gas mixture is 0.854.

6 0

3 years ago

What carbon atoms can bond with

elena-s [515]

Carbon has four valence electrons, so it can achieve a full outer energy level by forming four covalent bonds. When it bonds only with hydrogen, it formscompounds called hydrocarbons. Carbon can form single, double, or triple covalent bonds with other carbon atoms.

8 0

3 years ago

CAN SOMEONE PLEASE HELP?

IgorC [24]

Hey there!:

The 1s, 2s and 2p subshells are completely filled (a maximum of two electrons go into the 1s subshell and a maximum of two electrons go into the 2s subshell. The 2p subshell includes 3 orbitals, with 2 electrons maximum per orbital). The 3s subshell has only one of a maximum of two electrons.

Hope that helps!

4 0

4 years ago