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2 years ago

If you start with 30.0 grams of sodium fluoride, how many grams of magnesium fluoride will be produced?

Chemistry

1 answer:

Cloud [144]2 years ago

7 0

Answer:

22.2 g of MgF₂

Explanation:

We star from the reaction:

NaF → sodium fluoride

Mg → Magnesium

2NaF + Mg → MgF₂ + 2Na

2 moles of sodium fluoride react to 1 mol of Mg in order to produce 1 mol of magnesium fluoride and 2 moles of sodium.

We convert mass of the reactant to moles:

30 g . 1mol / 41.98 g = 0.715 mol

We assume that Mg is in excess

As ratio is 2:1, if we have 0.715 moles of NaF we may produce the half of moles, of MgF₂

0.715 mol /2 = 0.357 moles of MgF₂

We convert moles to mass: 0.357 mol . 62.30g /mol = 22.2 g

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2Al₂O₃ = 4Al + 3O₂

M(A₂O₃)=101.96 g/mol
m(Al₂O₃)=250 g

n(O₂)=3m(Al₂O₃)/{2M(Al₂O₃)}

n(O₂)=3*250/{2*101.96}=3.678 mol

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3 years ago

Choose the correct word that describes the description. If you add too much sugar to the water, some of the crystals will sit on

Triss [41]

Answer:

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Explanation:

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2 years ago

09. Where does Mitosis take place? I

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Answer:

In plants mitosis takes place throughout life in growing regions called the meristems. Replacements as cells wear out. The cells of the skin and bone marrow are sites of active mitosis replacing skin cells and red blood cells that only have a limited life. Repair.

Explanation:

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2 years ago

What is the percent lithium in lithium nitrate?

seropon [69]

Answer:

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Explanation:

4 0

3 years ago

A chemical reaction was used to produce 2.95 moles of copper(II) bicarbonate, Cu(HCO3)2.

BARSIC [14]

Answer:

About 547 grams.

Explanation:

We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.

To do so, we can use the initial value and convert it to grams using the molar mass.

Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:

$\displaystyle \begin{aligned} \text{MM}_\text{Cu(HCO$_3$)$_2$} &= (63.55 + 2(1.01)+2(12.01)+6(16.00))\text{ g/mol} \\ \\ &=185.59\text{ g/mol} \end{aligned}$

Dimensional Analysis:

$\displaystyle 2.95\text{ mol Cu(HCO$_3$)$_2$}\cdot \frac{185.59 \text{ g Cu(HCO$_3$)$_2$}}{1 \text{ mol Cu(HCO$_3$)$_2$}} \Rightarrow 547 \text{ g Cu(HCO$_3$)$_2$ }$

In conclusion, about 547 grams of copper (II) bicarbonate is produced.

8 0

3 years ago