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2 years ago

How many cobalt atoms are in 2 moles of cobalt?

1 answer:

4 0

That means that one mole of cobalt weighs 58.9332 grams (58.9332 g/mol). Based on that information, to convert 2 moles of cobalt to grams, we multiply 2 moles of cobalt by 58.9332.

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How does the air on a hot day compare with the air on a cold day?

MrMuchimi

Air on a hot day translates to hot air and air on a cold day translates to cold air. The hot air usually have more energy due to particles being more excited by the heat.

7 0

2 years ago

A compound has the empirical formula CH2O. If the compound’s molecular mass is 180 g/mol, determine the molecular formula of the

Umnica [9.8K]

The molecular formula will be a multiple of the empirical CH2O. One unit of CH2O has a mass of 12+2*1+16 = 30 g. This means that if our compound has a molecular mass of 180 g/mol, we can divide 180 / 30 = 6 units, and our compound has 6 units of CH2O. This means that its molecular formula is C6H12O6.

3 0

2 years ago

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0

3 years ago

An impurity sometimes found in Ca₃(PO₄)₂ is Fe₂O₃, which is removed during the production of phosphorus as ferrophosphorus (Fe₂P

Citrus2011 [14]

This impurity is troubling from an economic standpoint because it lead to decrease in the yield of phosphorus

Ferrophosphorus is a byproduct of phosphorus production in submerged-arc furnaces , by their reduction with carbon. It is formed from the iron oxide impurities.

Iron impurities present in the calcium phosphate will be precipitated out as the iron phosphate which eventually will lead to the decrease in the yield of phosphorous during the production of phosphorous.

Thus we can conclude that Fe₂P causes decrease in yield

Learn more about production of phosphorus at brainly.com/question/13337198

#SPJ4

6 0

1 year ago

What emotional state is important to maintain while reporting an accident in the lab?

sergeinik [125]

To keep calm and not go crazy

6 0

3 years ago

Read 2 more answers