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Dimas [21]
3 years ago
9

What product(s) are formed during this chemical reaction? (Will be two answers)

Chemistry
1 answer:
lianna [129]3 years ago
3 0

Answer:

BK. BK. BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK

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The equations given in the problem introduction can be added together to give the following reaction: overall: N2O4→2NO2 However
Murljashka [212]

Answer:

Reverse the 2 NO_2 \longrightarrow 2 NO + O_2  reaction

Explanation:

Reactions:

2 NO_2 \longrightarrow 2 NO + O_2

N_2O_4 \longrightarrow 2 NO + O_2

Overall:

N_2O_4 \longrightarrow 2 NO_2

As can be seen, in the overall reaction we have N_2O_4 in the reactants like in the second reaction and NO_2 in the products. The NO_2 is in the first reaction but as a reactant so we need to reverse that reaction:

2 NO + O_2 \longrightarrow 2 NO_2

N_2O_4 \longrightarrow 2 NO + O_2

Combining:

N_2O_4 + 2 NO + O_2\longrightarrow 2 NO + O_2 + 2 NO_2

N_2O_4 \longrightarrow 2 NO_2

4 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
Calculate the amount of grams of water that was used when 5525J of energy was used to boil this amount of water.
djverab [1.8K]

Answer:

2.445 g

Explanation:

Step 1: Given and required data

  • Energy in the form of heat required to boil the water (Q): 5525 J
  • Latent heat of vaporization of water (∆H°vap): 2260 J/g
  • Mass of water (m): ?

Step 2: Calculate the mass of water

We will use the following expression.

Q = ∆H°vap × m

m = Q / ∆H°vap

m = 5525 J / (2260 J/g)

m = 2.445 g

4 0
3 years ago
Rain washing away soil from a hillside
Aneli [31]

Answer:

deposition

Explanation:

8 0
3 years ago
How many valence electrons does lithium (Li) have available for bonding?
Marina CMI [18]
Lithium has 1 valence electron available for bonding. So its A.
6 0
3 years ago
Read 2 more answers
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