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Neporo4naja [7]
3 years ago
14

Which element is classified as a noble gas at STP?(1) hydrogen (3) neon(2) oxygen (4) nitrogen

Chemistry
1 answer:
strojnjashka [21]3 years ago
3 0
Neon is the element that is classified as a noble gas at STP.
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Type the correct answer in the box. Express your answer to three significant figures. This balanced equation shows the reaction
Ratling [72]

Answer:

514.5 g.

Explanation:

  • The balanced equation of the reaction is: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O.
  • It is clear that every 2.0 moles of NaOH react with 1.0 mole of H₂SO₄ to produce 1.0 mole of Na₂SO₄ and 2.0 moles of 2H₂O.
  • Since NaOH is in excess, so H₂SO₄  is the limiting reactant.
  • We need to calculate the no. of moles of 355.0 g of H₂SO₄:

n of H₂SO₄ = mass/molar mass = (355.0 g)/(98.0 g/mol) = 3.622 mol.

Using cross multiplication:

∵ 1.0 mol H₂SO₄ produces → 1.0 mol of Na₂SO₄.

∴ 3.622 mol H₂SO₄ produces → 3.662 mol of Na₂SO₄.

  • Now, we can get the theoretical mass of Na₂SO₄:

∴ mass of Na₂SO₄ =  no. of moles x molar mass = (3.662 mol)(142.04 g/mol) = 514.5 g.

8 0
3 years ago
Cu+2AgNO
baherus [9]

Answer:

Mass of Ag produced = 64.6 g

Note: the question is, how many grams of Ag is produced from 19.0 g of Cu and 125 g of AgNO3

Explanation:

Equation of the reaction:

Cu + 2AgNO3 ---> 2Ag + Cu(NO3)2

From the equation above, 1 mole of Cu reacts with 2 moles of AgNO3 to produce 2 moles of Ag and 1 mole of Cu(NO3)2.

Molar mass of the reactants and products are; Cu = 63.5 g/mol, Ag = 108 g/mol, AgNO3 = 170 g/mol, Cu(NO3)2 = 187.5 g/mol

To determine, the limiting reactant;

63.5 g of Cu reacts with 170 * 2 g of AgNO3,

19 g of Cu will react with (340 * 19)/63.5 g of AgNO3 =101.7 g of AgNO3.

Since there are 125 g of AgNO3 available for reaction, it is in excess and Cu is the limiting reactant.

63.5 g of Cu reacts to produce 108 * 2 g of Ag,

19 g of Cu will react to produce (216 * 19)/63.5 g of Ag = 64.6 g of Ag.

Therefore mass of Ag produced = 64.6g

6 0
3 years ago
20.352 mL of chlorine under a pressure of 680. mm Hg are
Lunna [17]

Answer:

0.01144L or 1.144x10^-2L

Explanation:

Data obtained from the question include:

V1 (initial volume) = 20.352 mL

P1 (initial pressure) = 680mmHg

P2 (final pressure) = 1210mmHg

V2 (final volume) =.?

Using the Boyle's law equation P1V1 = P2V2, the volume of the container can be obtained as follow:

P1V1 = P2V2

680 x 20.352 = 1210 x V2

Divide both side by 1210

V2 = (680 x 20.352)/1210

V2 = 11.44mL

Now we need to convert 11.44mL to L in order to obtain the desired result. This is illustrated below:

1000mL = 1 L

11.44mL = 11.44/1000 = 0.01144L

Therefore the volume of the container is 0.01144L or 1.144x10^-2L

7 0
3 years ago
Read 2 more answers
What is the charge of oxygen in Na2O?
Nuetrik [128]

In Na2O, what is the oxidation state of oxygen? In Na2O oxidation state of Na is 1+
3 0
2 years ago
Read 2 more answers
Fill in the blanks with the correct words to complete the sentence.
navik [9.2K]

Answer:

Explanation:

Salt water intrusion can cause the <u><em>fresh</em></u> water in wells to become contaminated with<u><em> salt</em></u>water.

4 0
3 years ago
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