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nlexa [21]
3 years ago
13

Calculate the molar mass of a solute in a 1.30 L solution with of 0.50 M

Chemistry
1 answer:
Veronika [31]3 years ago
6 0

Answer: 16 g/mol

Explanation: Molarity is Moles/Liters or Mol/L

(0.50 mol/L)(1.30L) = 0.65mol

(cancels out L)

Molar mass is g/mol so...

10.5g/0.65mol = 16.1 g/mol

Rounded to 2 sig figs is 16 g/mol

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Using the balanced equation for the combustion of ethane: 2C2H6 + 7O2 → 4CO2 + 6H2O, how many moles of O2 needed to produce 12 m
pickupchik [31]

Answer:

14 moles of oxygen needed to produce 12 moles of H2O.

Explanation:

We are given that balance eqaution

2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O

We have to find number of moles of O2 needed  to produce 12 moles of H2O.

From given equation

We can see that

6 moles of   H2O produced by Oxygen =7 moles

1 mole of   H2O produced by Oxygen=\frac{7}{6}moles

12 moles of H2O produced by Oxygen=\frac{7}{6}\times 12moles

12 moles of H2O produced by Oxygen=7\times 2moles

12 moles of H2O produced by Oxygen=14 moles

Hence, 14 moles of oxygen needed to produce 12 moles of H2O.

3 0
3 years ago
If you are 5 foot 10 inches, how tall are you in meters? Meters ( write you answer 4 significant figures e.g. 2.852)
horsena [70]
5 foot 10 inch = 1.55448 meter
3 0
3 years ago
1. What is the first element on the periodic table?
Elena L [17]
Hydrogen as it has only 1 proton and therefore only an atomic mass of 1...
7 0
3 years ago
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I wrote that all matter consists of tiny particles called atoms. I also thought all atoms of a specific
Westkost [7]

Answer:

John Dalton

Explanation:

John Dalton in 1808 suggested that all matter consists of tiny particles called atoms and that the atoms of a specific element are identical.

He postulated the Dalton's atomic theory which has the following important parts;

  • All matters consists of indivisible particles called atoms
  • Atoms of the same element are similar and are different from atoms of other elements.
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How do you calculate mass using density and volume?
vazorg [7]

D = M/v
M = Dv
V = M/D
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