Answer : The correct option is, (4) 6.0 mol
Explanation :
The given balanced chemical equation is,

In this reaction, lead undergoes reduction and chromium undergoes oxidation.
Oxidation reaction : It is the reaction in which a substance looses its electrons. In this oxidation state increases.
Reduction reaction : It is the reaction in which a substance gains electrons. In this oxidation state decreases.
Half reactions of oxidation-reduction are :
Oxidation : 
Reduction : 
From the reduction reaction, we conclude that 6 moles of electrons gained by the 3 moles of lead ions.
Hence, the correct option is, (4) 6.0 mole
Answer:
The answer is 1.778*10^-9
Explanation:
Using the formula
pH = -log[H^+]
Since the pH of the solution has been given to be 8.75,
Therefore,
8.75= -log [H^+]
Let's take H^+ to be x
8.75= -log x
Let's look for x, since the log is in base 10
X = 10^-8.75
X = 1.778 * 10^ -9
Therefore, H^+ =1.778 *10^-9
Answer:
Empirical formula of compound is C₄H₈O
Explanation:
Given data:
Mass of compound = 5.60 g
Mass of CO₂ = 13.7 g
Mass of H₂O = 5.60 g
Empirical formula of compound = ?
Solution:
Percentage of C:
13.7 g/5.60 g × 12/44× 100
2.45×0.273× 100 = 66.9%
Percentage of H:
5.60 g/ 5.60 g × 2.016/18 × 100
11.2%
Percentage of O:
(66.9% + 11.2%) - 100 = 21.9%
Grams atom of C , H, O
66.9/12 = 5.6
11.2 / 1.008 = 11.11
21.9 / 16 = 1.4
Atomic ratio:
C : H : O
5.6/1.4 : 11.11/1.4 : 1.4/1.4
4 : 8 : 1
Empirical formula:
C₄H₈O
It doesn't have enough solute. So say someone likes sweet tea. If you were to add more sugar it would still dissolve no matter how much is in the glass. Sorry if this is confusing.
This problem is being solved using Ideal Gas Equation.
PV = nRT
Data Given:
Initial Temperature = T₁ = 27 °C = 300 K
Initial Pressure = P₁ = constant
Initial Volume = V₁ = 8 L
Final Temperature = T₂ = 78 °C = 351 K
Final Pressure = P₂ = constant
Final Volume = V₂ = ?
As,
Gas constant R and Pressures are constant, so, Ideal gas equation can be written as,
V₁ / T₁ = V₂ / T₂
Solving for V₂,
V₂ = (V₁ × T₂) ÷ T₁
Putting Values,
V₂ = (8 L × 351 K) ÷ 300 K
V₂ = 9.38 L