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2 years ago

What is the percent by mass of 110.6 g of potassium fluoride (NaF) dissolved in 1500.0 g of

Chemistry

1 answer:

sveta [45]2 years ago

8 0

The percent by mass of 110.6 g of potassium fluoride (NaF) dissolved in 1500.0 g of water is 0.073733 %.

<h3>What is the mass percentage?</h3>

The mass percent of a solution is defined as the ratio of the mass of solute that is present in a solution, relative to the mass of the solution, as a whole.

$\rm\;Mass \;percentage =\frac{\;Mass \;of \;solute}{\;Mass \;of \;solution}$

Calculate the mass percent of potassium fluoride, using the formula:

$\rm\;Mass \;percentage =\frac{\;Potassium \;fluoride}{\;Mass \;of \;Water}$

$\rm\;Mass \;percentage =\frac{110.6}{1500.0}$

Mass percentage = 0.073733 %

Hence, the percent by mass of 110.6 g of potassium fluoride (NaF) dissolved in 1500.0 g of water is 0.073733 %.

Learn more about the mass percentage here;

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does a reaction occur when an aqueous solution of calcium iodide and sodium sulfate are combined. If a reaction does occur write

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Answer:

Yes. Ca²⁺(aq) + SO₄²⁻(aq) ⟶ CaSO₄(s)

Step-by-step explanation:

The possible reaction is

CaI₂ + Na₂SO₄ ⟶ CaSO₄ + 2NaI

To predict a reaction, you must know the solubility rules.

The important ones for this question are

Salts containing Na⁺ are soluble. Thus, NaI is soluble.
Most sulfates are soluble, but CaSO₄ is not. Thus, CaSO₄ is a precipitate, and a reaction occurs.

Molecular equation

CaI₂(aq) + Na₂SO₄(aq) ⟶ CaSO₄(s) + 2NaI(aq)

Ionic equation

Ca²⁺(aq) + 2I⁻(aq) + 2Na²⁺(aq) + SO₄²⁻(aq) ⟶ CaSO₄(s) + 2Na⁺(aq) + 2I⁻(aq)

Net ionic equation

Cancel all ions that appear on both sides of the reaction arrow (underlined).

Ca²⁺(aq) + 2I⁻(aq) + 2Na²⁺(aq) + SO₄²⁻(aq) ⟶ CaSO₄(s) + 2Na⁺(aq) + 2I⁻(aq)

The net ionic equation is

Ca²⁺(aq) + SO₄²⁻(aq) ⟶ CaSO₄(s)

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2 years ago

Please help me out on this question!!

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Answer:

Water H2O

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3 years ago

Consider the reaction Mg₂Si(s) + 4 H₂O(ℓ) → 2 Mg(OH)₂(aq) + SiH₄(g). How many grams of silane gas (SiH₄) are formed if 25.0 g of

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Answer:

10.60 grams of silane gas are formed.

Explanation:

From the reaction:

Mg₂Si(s) + 4H₂O(l) → 2Mg(OH)₂(aq) + SiH₄(g)

We know that the limiting reactant is Mg₂Si, so to find the mass of SiH₄ formed we need to calculate the number of moles of Mg₂Si:

$\eta_{Mg_{2}Si} = \frac{m_{Mg_{2}Si}}{M_{Mg_{2}Si}}$

Where:

m: is the mass of Mg₂Si = 25.0 g

M: is the molar mass of Mg₂Si = 76.69 g/mol

$\eta_{Mg_{2}Si} = \frac{m_{Mg_{2}Si}}{M_{Mg_{2}Si}} = \frac{25.0 g}{76.69 g/mol} = 0.33 moles$

Now, the stoichiometric relation between Mg₂Si and SiH₄ is 1:1 so:

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$m_{SiH_{4}} = \eta_{SiH_{4}}*M_{SiH_{4}} = 0.33 moles*32.12 g/mol = 10.60 g$

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