Answer:
The pressure, when the volume is reduced to 7.88L, is 846 torr (option A)
Explanation:
Step 1: Data given
The temperature of a gas = 25.0°C
AT 25 °C the gas occupies a volume of 10.0L and a pressure of 667 torr.
The volume reduces to 7.88 L but the temperature stays constant.
Step 2: Boyle's law
(P1*V1)/T1 = (P2*V2)/T2
⇒ Since the temperature stays constant, we can simplify to:
P1*V1 = P2*V2
⇒ with P1 = the initial pressure 667 torr
⇒ with V1 = the initial volume = 10.0 L
⇒ with P2 = the final pressure = TO BE DETERMINED
⇒ with V2 = the final volume = 7.88L
P2 = (P1*V1)/V2
P2 = (667*10.0)/7.88
P2 = 846 torr
The pressure, when the volume is reduced to 7.88L, is 846 torr (option A)
Explanation :
(a) 
This reaction is combustion reaction in which an oxygen react with a molecule to give its corresponding oxides ans water molecule.
(b) 
This reaction is a redox reaction or oxidation-reduction reaction in which sulfur get oxidized and oxygen get reduced.
(c) 
This reaction is a combination reaction in which the two reactants molecule combine to form a large molecule or product.
(d) 
This reaction is a decomposition reaction in which a large molecule or reactant decomposes to give two or more molecule or products.
(e) 
This reaction is a double displacement reaction in which the cation of two reactants molecule exchange their places to give two different products.
(f) 
This reaction is a combination reaction in which the two reactants combine to form a large molecule or product.
(g) 
This reaction is a double displacement reaction in which the cation of two reactants molecule exchange their places to give two different products.
(h) 
This reaction is combustion reaction in which a hydrocarbon react with an oxygen to give carbon dioxide and water as a products.
Answer:
pH of HNO₃ having an hydrogen ion concentration of 0.71M is 0.149
Explanation:
HNO₃ (aqueous) ⇄ H⁺ + NO3⁻
The pH is defined as the negative log of the hydrogen ion concentration
pH = - log [H⁺]
From the question, the hydrogen ion concentration is given as 0.71M, therefore
pH = -log [0.71]
= 0.149
Answer: 10.9 mol.
Explanation:
- To understand how to solve this problem, we must mention the reaction equation where water produced from PbO₂.
Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O
- Now, it is a stichiometric oriented problem, that 1 mole of PbO₂ produces 2 moles of H₂O.
Using cross multiplication:
1.0 mole of PbO₂ → 2.0 moles of H₂O
5.43 moles of PbO₂ → ??? moles of water
The moles of water produced = (5.43 x 2.0) = 10.86 moles ≅ 10.9 moles.
