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Sonja [21]
3 years ago
7

Calculate the molarity of a NaOH solution that is prepared by diluting 100mL of 0.20M NaOH with 150mL of H2O.

Chemistry
1 answer:
lianna [129]3 years ago
3 0

Molarity of solution = 0.08 M

<h3>Further explanation  </h3>

Molarity shows the number of moles of solute in every 1 liter of solution or mmol in each ml of solution  

\tt M=\dfrac{n}{V}  

Where  

M = Molarity  

n = number of moles of solute  

V = Volume of solution  

  • mol of NaOH

\tt mol=0.2~M\times 100~ml=20~mlmol=0.02~mol

  • molarity

Volume of solution = 100 ml + 150 ml = 250 ml

\tt M=\dfrac{0.02~mol}{0.250~L}=0.08

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A gas system has an initial number of moles of 0.693 moles with the volume unknown. When the number of moles changes to 0.928 mo
lara [203]

Answer:

The initial volume in mL is 5959.2 mL

Explanation:

As the number of moles of a gas increases, the volume also increases.  Hence, number of moles and volumes are directly proportional i.e

n ∝ V

Where n is the number of moles and V is the volume

Then, n = cV

c is the proportionality constant

∴n/V = c

Hence n₁/V₁ = n₂/V₂

Where n₁ is the initial number of moles

V₁ is the initial volume

n₂ is the final number of moles

and V₂ is the final volume.

From the question,

n₁ = 0.693 moles

V₁ = ?

n₂ = 0.928 moles

V₂ = 7.98 L

Putting the values into the equation

n₁/V₁ = n₂/V₂

0.693 / V₁ = 0.928 / 7.98

Cross multiply

∴ 0.928V₁ = 0.693 × 7.98  

0.928V₁ = 5.53014

V₁ = 5.53014/0.928

V₁ = 5.9592 L

To convert to mL, multiply by 1000

∴ V₁ = 5.9592 × 1000 mL

V₁ = 5959.2 mL

Hence, the initial volume in mL is 5959.2 mL

5 0
3 years ago
When nahco3 completely decomposes, it can follow this balanced chemical equation: 2nahco3 → na2co3 h2co3 determine the theoretic
BigorU [14]

Theoretical yield = 2.397

The product could be sodium carbonate

percent yield = 98.456%

When nahco3 completely decomposes, it can follow this balanced chemical equation:

2nahco3 → na2co3 h2co3

If the mass of the NaHCO3 sample is 3.80 g, we must use stoichiometry to calculate the theoretical yields of each of the products.

mass of NaHCO₃ = 3.80 g

molar mass of NaHCO₃ = 84 g/mol

so the no of moles of NaHCO₃ = 3.80/84 =  0.0452 mol

You see, one mole of sodium carbonate and one mole of hydrogen carbonate are produced from two moles of sodium bicarbonate.

so, the no of moles of sodium carbonate = 0.0452/2 = 0.0226 mol

∴ mass of sodium carbonate ( Na₂CO₃) = no of moles of Na₂CO₃ × molar mass of Na₂CO₃

=  0.0226 × 106 ≈ 2.397 g

no of moles of hydrogen carbonate = 0.0452/2 = 0.0226 mol

mass of the hydrogen carbonate ( H₂CO₃) = no of moles of H₂CO₃ × molar mass of H₂CO₃

= 0.0226 × 62 g = 1.401 g

mass of one of the products was measured to be 2.36 g , from above data, we can say it must be sodium carbonate because value is the nearest of 2.397 g.

percentage yield = experimental yield/theoretical yield × 100

here experimental yield of Na₂CO₃ = 2.36 g

and theoretical yield of Na₂CO₃ = 2.397 g

∴ % yield = 2.36/2.397 × 100 ≈ 98.456%

Therefore the percentage yield of the product is 98.456%

To learn more about percentage yield visit:

brainly.com/question/22257659

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Answer:

C

Explanation;

Maybe is C

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balandron [24]

Answer:

<em>At equilibrium, the rate of the forward, and the reverse reactions are equal.</em>

Explanation:

In an equilibrium chemical reaction, the rate of forward reaction, is equal to the rate of reverse reaction. Note that the reactions does not cease at equilibrium, but rather, the reactants are converted to product, at the same rate at which the product is also being converted into the reactants in the reaction. When chemical equilibrium is reached, a careful calculation of the value of equilibrium constant is approximately equal to 1.

NB: If the value of equilibrium constant is far far greater than 1, then the reaction will favors more of the forward reaction, and if far far less than 1, the reaction will favor more of the reverse reaction.

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