Water is the answer hope this helps
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
A grounding electrode is any object that directly links to the earth. They are most times used to divert electricity from the elements.
- Swimming pool structures and structural <u>reinforcing steel. 250.52(B)(3)</u><u>,</u> [680.26(B)(1), and (B)(2)] shall not be used as a grounding electrode.
In code 250.52(B)(3) it is clearly specified that the bonding grid and reinforcing steel that is related to a pool should not be used as grounding electrodes.
This is essential because when a metal that lies beneath a swimming pool is used as a grounding electrode, current from nearby electrical systems can be introduced into the pool.
This could cause the electrocution of anybody in the swimming pool at that time.
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Answer:
1.33 L.
Explanation:
- We can use the general law of ideal gas: PV = nRT.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and T are constant, and have different values of P and V:
<em>(P₁V₁) = (P₂V₂)</em>
<em></em>
Knowing that:
V₁ = 4.0 L, P₁ = 2.0 atm,
V₂ = ??? L, P₂ = 6.0 atm.
- Applying in the above equation
(P ₁V₁) = (P₂V₂)
<em>∴ V₂ = P ₁V₁/P₂</em> = (2.0 atm)(4.0 L)/(6.0 atm) =<em> 1.33 L.</em>
Answer: The final volume of this solution is 0.204 L.
Explanation:
Given: Molarity of solution = 2.2 M
Moles of solute = 0.45 mol
Molarity is the number of moles of solute present divided by volume in liters.
Substitute the values into above formula as follows.
Thus, we can conclude that the final volume of this solution is 0.204 L.
