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3 years ago

Human activities can accelerate the process of

Chemistry

1 answer:

Aleks [24]3 years ago

4 0

I believe the answer is C) both

Hope this helps

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Answer:

carbohydrates

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3 years ago

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How many atoms in the pictured molecule can form hydrogen bonds with water molecules anatomy?

garri49 [273]

55..............................

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3 years ago

Is electrical conductivity an extensive property or an intensive property?

Thepotemich [5.8K]

Your Answer Will Be Intensive Property

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3 years ago

Consider the reaction between reactants s and o2: 2s(s)+3o2(g)→2so3(g)if a reaction vessel initially contains 7 mols and 9 mol o

nadya68 [22]

Answer: -

1 mol

Explanation: -

Number of moles of Sulphur S = 7

Number of moles of O2 = 9

The balanced chemical equation for the reaction is

2S (s)+3 O2 (g)→2SO3(g)

From the above reaction we can see that

3 mol of O2 react with 2 mol of S

9 mol of O2 will react with $\frac{2 mol S x 9 mol O2}{3 mol O2}$

= 6 mol of S

Unreacted S = 7 - = 1 mol.

If a reaction vessel initially contains 7 mol S and 9 mol O2

1 mole of s will be in the reaction vessel once the reactants have reacted as much as possible

6 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