What is the balance equation for HgO(s)- Hg(l)+O2(g)

(a) What human activity is responsible for the formation of smog and acid rain? (b) What kinds of harm does photochemical smog c

Lunna [17]

but u have a good person to talk about that is a bit of an act like I swear to God that

Explanation:

can you send the invoice to the b account on your website so I just put in a sec and I can send it here and send you the info for my credit card

6 0

3 years ago

Do all multicellular organisms follow the same pattern of organization?

natta225 [31]

So so sorry if i'm wrong but i'm pretty sure no

3 0

3 years ago

Write down the molecular formula and molecular weight of carbon dioxide .

Finger [1]

Answer:

CO2

Molar mass: 44.01 g/mol

Explanation:

CO2

3 0

3 years ago

You carefully weigh out 20.00 g of CaCO3 powder and add it to 81.00 g of HCl solution. You notice bubbles as a reaction takes pl

Zepler [3.9K]

Answer:

The relevant equation is:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Explanation:

1 mol of calcium carbonate can react to 2 moles of Hydrochloric acid to produce 1 mol of water, 1 mol of calcium chloride and 1 mol of carbon dioxide.

The formed CO₂ is the reason why you noticed bubbles as the reaction took place

3 0

3 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