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densk [106]
4 years ago
7

Making solutions is an extremely important component to real-life chemistry. If you make 3.00 L of a solution using 90.0 g of so

dium hydroxide, what is the final concentration?
Chemistry
1 answer:
kumpel [21]4 years ago
5 0

Answer:

Final concentration of NaOH = 0.75 M

Explanation:

For NaOH :-

Given mass = 90.0 g

Molar mass of NaOH = 39.997 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{90.0\ g}{39.997\ g/mol}

Moles\ of\ NaOH= 2.2502\ mol

Molarity is defined as the number of moles present in one liter of the solution. It is basically the ratio of the moles of the solute to the liters of the solution.

The expression for the molarity, according to its definition is shown below as:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Where, Volume must be in Liter.

It is denoted by M.

Given, Volume = 3.00 L

So,

Molarity=\frac{2.2502\ mol}{3.00\ L}=0.75\ M

<u>Final concentration of NaOH = 0.75 M</u>

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consider the reaction between calcium oxide and carbon dioxide: cao ( s ) + co 2 ( g ) → caco 3 ( s ) a chemist allows 14.4 g of
MakcuM [25]

Answer:

CaO is the limiting reagent

Theoritical yield = 25.71 g

% Yield = 75.44%

Explanation:

1 mole = Molar mass of the substance

Molar Mass of CaO = 56 g/mol

Molar Mass of CaCO3 = 100 g/mol

Molar mass of CO2 = 44 g/mol

The balanced Equation is :

CaO + CO_{2}\rightarrow CaCO_{3}

1 mole of CaO reacts with = 1 mole of CO2

56 g of CaO reacts with = 44 g of CO2

1 g of CaO reacts with =

\frac{44}{56}

= 0.785 g of CO2

So,

<u>14.4 g of CaO</u><u> </u>must react with = (14.4 x 0.785) g of CO2

= 11.31 g of CO2

<u>Needed = 11.31 g</u>

<u>Available CO2  = 13.8 g</u><u> </u>(given)

So CO2 is in excess , hence<u> CaO is the limiting reagent and product will produce from 14.4 g of CaO</u>

CaO + CO_{2}\rightarrow CaCO_{3}

1 mole of CaO will produce 1 mole pf CaCO3

56 g of CaO produce = 100 g of CaCO3

1 g  of CaO produce =

\frac{100}{56}

= 1.785 g of CaCO3

14.4 g of CaO will produce = (1.785 x 14.4) g of CaCO3

= 25.71 g of CaCO3

Theoritical Yield of CaCO3 = 25.71 g

Actual yield = 19.4 g

Percent Yield =

\frac{Actual\ yield}{Theoritical\ yield}\times 100

\frac{19.4}{25.71}\times 100

= 75.44 %

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3 years ago
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Answer:

Explanation:

Its letstute.

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Explanation:

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Cells have many structure inside of them called organelles which allow it to function and survive

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