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Dimas [21]
3 years ago
13

3.95 g of sugar (C6H12O6) is dissolved in water to make 158 mL of solution. Find the molarity.

Chemistry
1 answer:
12345 [234]3 years ago
8 0

Answer:

[C₆H₁₂O₆] = 0.139 M

Explanation:

Molarity si defined as a sort of concentration. It indicates the moles of solute that are contained in 1 L of solution.

We can also say, that molarity are the mmoles of solute contained in 1 mL of solution.

For this case, the solute is sugar (glucose). Let's determine M (mmol/mL)

(3.95 g . 1mol / 180g) . (1000 mmol / 1mol) / 158 mL

We determine moles, we convert them to mmoles, we divide by mL

M = 0.139 M

Moles = 3.95 g . 1mol / 180g → 0.0219 mol

We convert mL to L → 158 mL . 1L/1000mL = 0.158L

M = 0.0219 mol / 0.158L = 0.139 M

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Water is a produced when 30.0 grams of hydrogen reacts with 80.0 grams of oxygen what is the limiting reagent
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The balanced equation for the reaction is as follows
2H₂ + O₂ --> 2H₂O
stoichiometry of H₂ to O₂ is 2:1
number of H₂ moles - 30.0 g / 2 g/mol = 15 mol 
number of O₂ moles - 80.0 g / 32 g/mol = 2.5 mol
limiting reactant is the reagent in which only a fraction is used up in the reaction
if H₂ is the limiting reactant 
if 2 mol of H₂ requires 1 mol of O₂
then 15 mol of H₂ requires 1/2 x 15.0 = 7.5 mol of O₂
but only 2.5 mol of O₂ is required 
this means that O₂ is the limiting reagentt and H₂ is in excess
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4 years ago
Unit processes in charge of smelting​
Svet_ta [14]

Answer:

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Hope I helped!

Explanation:

8 0
3 years ago
What is the number of electrons in the highest occupied energy level of an element in Group 5A?
nignag [31]

5 is the answer i took the test

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3 years ago
At a certain temperature and pressure, one liter of CO2 gas weighs 1.95 g.
AysviL [449]

Answer:

1.332 g.

Explanation:

  • We can use the general law of ideal gas: <em>PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

  • At the same T and P and constant V (1.0 L), different gases have the same no. of moles (n):

<em>∴ (n) of CO₂ = (n) of C₂H₆</em>

<em></em>

∵ n = mass/molar mass

<em>∴ (mass/molar mass) of CO₂ = (mass/molar mass) of C₂H₆</em>

mass of CO₂ = 1.95 g, molar mass of CO₂ = 44.01 g/mol.

mass of C₂H₆ = ??? g, molar mass of C₂H₆ = 30.07 g/mol.

<em>∴ mass of C₂H₆ = [(mass/molar mass) of CO₂]*(molar mass) of C₂H₆</em> = [(1.95 g / 44.01 g/mol)] * (30.07 g/mol) =<em> 1.332 g.</em>

<em></em>

7 0
4 years ago
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