You can be electrocuted if you try to use water to put out a class

A sample of strontium bicarbonate weighs 5.0020 mg. How many oxygen atoms are in the sample? I have the answer, but don't unders

hodyreva [135]

There are 14.4 * 10^18 oxygen atoms in 5.0020 mg of strontium bicarbonate.

Mass of strontium bicarbonate Sr(HCO3)2 = 5.0020 mg

Molar mass of strontium bicarbonate Sr(HCO3)2 = 209.6537 g/mol

Number of moles of strontium bicarbonate Sr(HCO3)2 = 5.0020 * 10^-3g/209.6537 g/mol

= $2.3858 * 10^-5$ moles

Given that we have 6 oxygen atoms per molecule of Sr(HCO3)2, the total number of oxygen atoms in Sr(HCO3)2 becomes;

$2.3858 * 10^-5 moles * 6.02 * 10^23 = 14.4 * 10^18 oxygen atoms$

Learn more: brainly.com/question/9743981

5 0

3 years ago

Which of the following is true for compounds?

soldi70 [24.7K]

Compound are formed by two or more elements chemically combined. For example: H^2O is the water formula, this is a compound because you two elements which are Hydrogen and Oxygen and together they form a compound. The (^2) is the amount of atoms the formula has, in this case Hydrogen has two atoms and Oxygen is neutral.

4 0

3 years ago

SiCl4(l) + H2O(l) → SiO2(s) + HCl(aq)

Sveta_85 [38]

I assume you’re looking for a balanced equation.
SiCl4 + 2H2O = SiO2 + 4HCl

5 0

4 years ago

Water comes from different a.levels b.phases c.sources d.stages

bezimeni [28]

Answer:

C. sources

Explanation:

6 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

You can be electrocuted if you try to use water to put out a class_ fire ?