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lidiya [134]
2 years ago
10

Read the given equation. Na2O2 + CO2 → Na2CO3 + O2 What volume of O2 gas is produced from 2.80 liters of CO2 at STP? 5.60 liters

4.20 liters 2.10 liters 1.40 liters
Chemistry
2 answers:
OleMash [197]2 years ago
8 0

The equation should be balanced as follows  2Na2O2 + 2CO2⇒ 2Na2CO3 +O2

2 moles of carbon (IV) oxide give 1 mole of oxygen.

Therefore, the mole ratio is of CO2 to O2 is 2:1

if 2.8 liters of CO2 were used, then the volume of O2 produced will be given by:   (2.80liters× 1)/2= 1.40 liters

umka21 [38]2 years ago
3 0

Answer:

The correct answer is 1.40 L.

Explanation:

Volume of CO_2 = 2.80 L

At STP. 1 mol of gas occupies = 22.4 L

Then 2.80 L of volume will be occupied by:

\frac{1}{22.4 L}\times 2.80 L=0.125 mol

2Na_2O_2 + 2CO_2 \rightarrow 2Na_2CO_3 + O_2

Acording to reaction 2 mol of CO_2 gives 1mol of O_2

Then 0.125 mol of CO_2 will give:

\frac{1}{2}\times 0.125 mol=0.0625 mol of O_2

Volume occupied by 0.0625 mol of oxygen gas at STP :

0.0625 \times 22.4 L= 1.40 L

1.40 L of oxygen gas will be produced.

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For O: atomic number = 8 # neutrons = 8

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3 years ago
What is velocity?
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A. how fast something moves in a specific direction
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2 years ago
Consider the following system at equilibrium:
Vika [28.1K]

Answer:

1) Rightward shift

2) Rightward shift

3) Leftward shift

4) Leftward shift

5) Leftward shift

6) Rightward shift

7) No shift

8) No shift                                                              

   

Explanation:

To evaluate each case we need to consider Le Chatelier's Principle, which states that the adding of additional reactants or products to a system will shift the equilibrium in the opposite direction, to maintain the equilibrium of the system. On the contrary, if we remove a reactant or a product in the system, the equilibrium will be shifted in the direction of the reactant or product reduced, to produce more of it (and thus maintain balance).        

Taking into account the above, let's see each statement, in the following equation:

A(aq) + B(aq)  ⇄  2C(aq)    (1)

1) Increase A. This will cause a rightward shift in equation 1 in order to consume the reactant added.

2) Increase B. Same as 1), this will cause a rightward in equation 1.

3) Increase C. This will cause a leftward shift in order to consume the excess of product in the system.  

4) Decrease A. This will produce a leftward shift to produce the reactant that is being reduced.

5) Decrease B. Same as 4), a leftward shift.

6) Decrease C. This will produce a rightward shift to produce the product that is being reduced.

7) Double A, half B. The double A will cause a rightward shift and the half B will produce a leftward shift, which results in no shift.

8) Double both B and C. Double B will produce a rightward shift and double C will produce the contrary, a leftward shift, so the final result is no shift.

               

I hope it helps you!

4 0
3 years ago
Iodine, I2, is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed. What is the temperature
Stels [109]
<h3>Answer:</h3>

382.63 K

<h3>Explanation:</h3>

We are given;

  • Volume of Iodine as 71.4 mL
  • Mass of Iodine as 0.276 g
  • Pressure of Iodine as 0.478 atm

We are required to calculate the temperature of Iodine

  • We are going to use the ideal gas equation;
  • According to the ideal gas equation; PV = nRT, where R is the ideal gas constant, 0.082057 L.atm/mol.K.
  • Rearranging the formula;

T = PV ÷ nR

But, n, the number of moles = Mass ÷ Molar mass

Molar mass of iodine = 253.8089 g/mol

Thus, n = 0.276 g ÷ 253.8089 g/mol

           = 0.001087 moles

Therefore;

T = (0.478 atm × 0.0714 L) ÷ (0.001087 moles × 0.082057)

  = 382.63 K

Thus, the temperature of Iodine in Kelvin is 382.63 K

3 0
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When Rutherford ran his gold foil experiment, he expected to see results like those in the Plum Pudding Atom simulation. Instead
DENIUS [597]

Rutherford's result can not be explained on the basis of the plum pudding model because of the fact that, since some alpha particles were deflected, the atom must contain a small region with a strong electric charge.

The empirical study of the atom led to the emergence of several models of the atom. In the Plum - pudding model, the atom was regarded as a positively charged sphere with embedded negative charges.

This model can not interpret the Rutherford experiment since alpha particles were deflected. The deflection of alpha particles means that, the atom must contain a small region with a strong electric charge.

Learn more: brainly.com/question/730256

4 0
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