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Alex
3 years ago
12

Perform the following mathematical operations and express the correct answer to the proper number of significant figures or deci

mal places as required.
7.71 + 4.31 + 1.7 + 4.00141=

7.71 x 4.31 x 1.7 x 4.00141=
Chemistry
1 answer:
nordsb [41]3 years ago
7 0

Answer: 7.71 + 4.31 + 1.7 + 4.00141= 17.72141

7.71 x 4.31 x 1.7 x 4.00141=  226.04433255

Explanation:

(i) 7.71 + 4.31 + 1.7 + 4.00141

= 7.71000+4.31000+1.70000+ 4.00141   [make like decimals]

= 17.72141    [By adding the corresponding places ]

(ii)  7.71 x 4.31 x 1.7 x 4.00141 =( 7.71 x 4.31) x 1.7 x 4.00141

= 33.2301 x 1.7 x 4.00141

= (33.2301 x 1.7) x 4.00141

=  56.49117 x 4.00141

= ( 56.49117 x 4.00141 )

=  226.04433255

Hence,  7.71 + 4.31 + 1.7 + 4.00141= 17.72141

7.71 x 4.31 x 1.7 x 4.00141=  226.04433255

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Part B When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 22.8 g of carbon were burned in the pres
aksik [14]

<u>Answer:</u> The amount of carbon dioxide gas produced in the reaction is 83.6 grams

<u>Explanation:</u>

As, some amount of oxygen gas is left after the reaction is completed. So, it is present in excess and is considered as an excess reagent.

Thus, carbon is considered as a limiting reagent because it limits the formation of product.

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

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

Given mass of carbon = 22.8 g

Molar mass of carbon = 12 g/mol

Putting values in equation 1, we get:

\text{Moles of carbon}=\frac{22.8g}{12g/mol}=1.9mol

The chemical equation for the reaction of carbon and oxygen gas follows:

C+O_2\rightarrow CO_2

By Stoichiometry of the reaction:

1 mole of carbon produces 1 mole of carbon dioxide gas

So, 1.9 moles of carbon will produce = \frac{1}{1}\times 1.9=1.9moles of carbon dioxide gas

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 1.9 moles

Putting values in equation 1, we get:

1.9mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(1.9mol\times 44g/mol)=83.6g

Hence, the amount of carbon dioxide gas produced in the reaction is 83.6 grams

6 0
3 years ago
Plz help me with this question plzzzzz​
LekaFEV [45]

Answer:

6

Explanation:

Formula: Al2O3

If we require 2Al2O3

                       We divide 2 by 3

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8 0
3 years ago
Using the equation below, if you have 4.3 mol of nitrogen tribromide and
ankoles [38]

Answer:

sodium hydroxide is the limiting reactant

Explanation:

The first step is usually to put down the balanced reaction equation. This is the first thing to do when solving any problem related to stoichiometry. The balanced reaction equation serves as a guide during the solution.

2NBr3 + 3NaOH = N2 + 3NaBr + 3HOBr

Let us pick nitrogen gas as our product of interest. Any of the reactants that gives a lower number of moles of nitrogen gas is the limiting reactant.

For nitrogen tribromide

From the balanced reaction equation;

2 moles of nitrogen tribromide yields 1 mole of nitrogen gas

4.3 moles of nitrogen tribromide will yield 4.3 ×1/ 2 = 2.15 moles of nitrogen gas

For sodium hydroxide;

3 moles of sodium hydroxide yields 1 mole of nitrogen gas

5.9 moles of sodium hydroxide yields 5.9 × 1/ 3= 1.97 moles of nitrogen gas

Therefore, sodium hydroxide is the limiting reactant.

8 0
3 years ago
The Bohr model of the atom explained why emission spectra are discrete. It could also be used to explain the photoelectric effec
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8 0
3 years ago
Read 2 more answers
A rigid tank contains 1.40 moles of an ideal gas. Determine the number of moles of gas that must be withdrawn from the tank to l
sergey [27]

Answer : The final number of moles of gas that withdrawn from the tank to lower the pressure of the gas must be, 0.301 mol.

Explanation :

As we know that:

PV=nRT

At constant volume and temperature of gas, the pressure will be directly proportional to the number of moles of gas.

The relation between  pressure and number of moles of gas will be:

\frac{P_1}{P_2}=\frac{n_1}{n_2}

where,

P_1 = initial pressure of gas = 24.5 atm

P_2 = final pressure of gas = 5.30 atm

n_1 = initial number of moles of gas = 1.40 moles

n_2 = final number of moles of gas = ?

Now put all the given values in the above expression, we get:

\frac{24.5atm}{5.30atm}=\frac{1.40mol}{n_2}

n_2=0.301mol

Therefore, the final number of moles of gas that withdrawn from the tank to lower the pressure of the gas must be, 0.301 mol.

8 0
3 years ago
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