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marshall27 [118]
3 years ago
5

H2O (s) ⇒ H2O What would be the best argument against writing this in the form of a chemical equation?

Chemistry
1 answer:
Goshia [24]3 years ago
5 0

Answer: Since this is merely a Physical Change, it is improper to write it as a chemical equation.

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3. In the formation of acid rain, sulfur dioxide reacts with oxygen and water in the air to form sulfuric acid. Write a balanced
Andreyy89

Answer:

SO₂ + 0.5 O₂ + H₂O → H₂SO₄

3.83 g

Explanation:

In the formation of acid rain, sulfur dioxide reacts with oxygen and water in the air to form sulfuric acid. The balanced chemical equation is:

SO₂ + 0.5 O₂ + H₂O → H₂SO₄

The molar mass of SO₂ is 64.07 g/mol. The moles of SO₂ corresponding to 2.50 g are:

2.50 g × (1 mol/64.07 g) = 0.0390 mol

The molar ratio of SO₂ to H₂SO₄ is 1:1. The moles of H₂SO₄ formed are 0.0390 moles.

The molar mass of H₂SO₄ is 98.08 g/mol. The mass of H₂SO₄ is:

0.0390 mol × 98.08 g/mol = 3.83 g

4 0
3 years ago
A chemical formula is shown the subscription in a chemical formula represent the
Alenkinab [10]

Answer:

A, option is the correct answer of this question

7 0
3 years ago
Four samples of gas each exert 675 mm Hg in separate 3.5 L containers. What pressure will they exert if they are all placed in a
DedPeter [7]
The pressure that  will  be   exerted if  four sample  of gas are placed in a single  3.5 container  is calculated as below

if  each gas occupies 675  mmhg

what about  4 gases in the sample
by cross  multiplication

= 675 mm hg x 4/1 = 2.7  x10^3mmhg (answer D)
3 0
3 years ago
What is the solubility of CaCl2 at 20°C
slavikrds [6]

Answer:1,01

Explanation:

3 0
2 years ago
Calculate the molarity of 13.1 g of NaCl in 727 mL of solution
valentina_108 [34]

Answer:

The molarity of the solution is 0,31 M

Explanation:

We calculate the weight of 1 mol of NaCl from the atomic weights of each element of the periodic table. Then, we calculate the molarity, which is a concentration measure that indicates the moles of solute (in this case NaCl) in 1000ml of solution (1 liter)

Weight 1 mol NaCl= Weight Na + Weight Cl= 23 g + 35, 5 g= 58, 5 g

58, 5 g-----1 mol NaCl

13,1 g ---------x= (13,1 g x 1 mol NaCl)/58, 5 g= 0, 224 mol NaCl

727 ml solution------ 0, 224 mol NaCl

1000ml solution------x= (1000ml solutionx0, 224 mol NaCl)/727 ml solution

x=0,308 mol NaCl---> <em>The solution is 0,31 molar (0,31 M)</em>

5 0
3 years ago
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