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ra1l [238]
3 years ago
6

What mass of ni has as many atoms as there are n atoms in 63.0 g of no2?

Chemistry
2 answers:
adoni [48]3 years ago
8 0

Answer : The mass of Ni is, 80.346 g

Solution :

First we have to calculate the number of moles.

\text{Moles of }NO_2=\frac{\text{Mass of }NO_2}{\text{Molar mass of }NO_2}=\frac{63g}{46g/mole}=1.369moles

Now we have to calculate the number of atoms of NO_2

As 1 moles NO_2 contains 6.022\times 10^{23} number of molecules

So, 1.369 moles NO_2 contains 1.369\times 6.022\times 10^{23} number of molecules

One NO_2 molecule has one nitrogen atom then 1.369\times 6.022\times 10^{23} number of molecules of NO_2 will contain 1.369\times 6.022\times 10^{23} number of nitrogen atoms

As per question number of Ni atoms is equal to the number of nitrogen atoms

Number of Ni atoms=1.369\times 6.022\times 10^{23} atoms

Number of moles = \frac{1.369\times 6.022\times 10^{23}}{6.022\times 10^{23}}=1.369moles

Mass of Nickel = Moles\times \text{Molecular mass}}

Molecular mass of Nickel = 58.69 g/mol

Mass of nickel = 1.369mol\times 58.69g/mol=80.346grams

djverab [1.8K]3 years ago
6 0

Answer:

80.42 g

Explanation:

NO2 has molar mass equal to 46 g/mol ( 14 of N + 2x16 of O). So, for 63.0 g of NO2, the number of moles are:

n = mass/molar mass

n = 63/46

n = 1.37 moles

By Avogadro's Law, 1 mol has 6.02x10^{23} atoms, for any substance. So, the substances will have the same number of atoms if they had the same number of moles

Then, n of Ni is 1.37 moles.

The molar mass of Ni is 58.7 g, so:

n = mass (m)/ molar mass

1.37 = m/58.7

m = 1.37x58.7

m = 80.42 g

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vodomira [7]

Answer:

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Explanation:

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2Na+O2-->2Na20

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The following equilibrium is formed when copper and bromide ions are placed in a solution:
JulsSmile [24]

Answer:

A)

1. Reaction will shift rightwards towards the products.

2. It will turn green.

3. The solution will be cooler..

B) It will turn green.

Explanation:

Hello,

In this case, for the stated equilibrium:

heat + Cu(H_2O)_6 ^{+2} (blue) + 4Br^- \rightleftharpoons 6H_2O + CuBr_4^{-2} (green)

In such a way, by thinking out the Le Chatelier's principle, we can answer to each question:

A)

1. If potassium bromide, which adds bromide ions, is added more reactant is being added to the solution, therefore, the reaction will shift rightwards towards the products.

2. The formation of the green complex is favored, therefore, it will turn green.

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B) In this case, as the heat is a reactant, if more heat is added, more products will be formed, which implies that it will turn green.

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For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi
natka813 [3]

<u>Answer:</u> The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       ....(1)

  • <u>For Iron:</u>

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:  

\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol

For the given chemical reaction:

2Fe(s)+O_2(g)\rightarrow 2FeO(s)

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = \frac{2}{2}\times 0.0771=0.0771mol of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:  

0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g

To calculate the percentage yield of iron (ii) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

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