Answer:
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
Explanation:
Hello,
In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.
Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.
Regards.
Explanation:
To solve this problem, follow these steps;
- Obtain a balanced equation of the reaction and familiarize with the reactants and products.
- Find the number of moles of the reacting species since they are the known matter in terms of quantity.
- From the number of moles, determine the limiting reactant.
- The limiting reactant is the one given in short supply.
- Such reactant determines the extent of the reaction.
- Compare the moles of this specie to that of the products using the balanced equation.
- Obtain the mole of the desired product and find the mass or desired quantity.
- simply work from the known specie to the unknown
learn more:
Number of moles brainly.com/question/13064292
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Answer:
248 mL
Explanation:
According to the law of conservation of energy, the sum of the heat absorbed by water (Qw) and the heat released by the coffee (Qc) is zero.
Qw + Qc = 0
Qw = -Qc [1]
We can calculate each heat using the following expression.
Q = c × m × ΔT
where,
- ΔT: change in the temperature
163 mL of coffee with a density of 0.997 g/mL have a mass of:
163 mL × 0.997 g/mL = 163 g
From [1]
Qw = -Qc
cw × mw × ΔTw = -cc × mc × ΔTc
mw × ΔTw = -mc × ΔTc
mw × (54.0°C-25.0°C) = -163 g × (54.0°C-97.9°C)
mw × 29.0°C = 163 g × 43.9°C
mw = 247 g
The volume corresponding to 247 g of water is:
247 g × (1 mL/0.997 g) = 248 mL
the most free particles are gases, then liquids, solids
