Answer:
The mass of 0.280 mole sample of sodium hydroxide NaOH is 11.2 grams.
Explanation:
To know the mass in grams of 0.280 moles of sample of sodium hydroxide NaOH, you must know the molar mass of the compound, that is, the mass of one mole of a substance, which can be an element or a compound.
So you know:
- Na: 23 g/mole
- O: 16 g/mole
- H: 1 g/mole
So, the molar mass of NaOH is:
NaOH= 23 g/mole + 16 g/mole+ 1 g/mole= 40 g/mole
Then the following rule of three can be applied: if in 1 mole of sodium hydroxide there are 40 grams, in 0.280 moles how much mass is there?
mass= 11.2 grams
<u><em>The mass of 0.280 mole sample of sodium hydroxide NaOH is 11.2 grams.</em></u>
Answer:
669.48 kJ
Explanation:
According to the question, we are required to determine the heat change involved.
We know that, heat change is given by the formula;
Heat change = Mass × change in temperature × Specific heat
In this case;
Change in temperature = Final temp - initial temp
= 99.7°C - 20°C
= 79.7° C
Mass of water is 2000 g ( 2000 mL × 1 g/mL)
Specific heat of water is 4.2 J/g°C
Therefore;
Heat change = 2000 g × 79.7 °C × 4.2 J/g°C
= 669,480 joules
But, 1 kJ = 1000 J
Therefore, heat change is 669.48 kJ
I think the answer is D no change. Though you add more CO2, but the pressure is not mentioned. If the pressure is constant and the reaction is already balanced, the H2O is also saturation and can not absorb more CO2.
Answer:
A
Explanation:
heat energy is transfered through a hot object touching a cold object
The actual yield is 43 g Cl₂.
The <em>limiting reactant was MnO₂</em> because it gave the smaller mass of Cl₂.
∴ The <em>theoretical yield</em> is 60.25 g Cl₂.
% yield = actual yield/theoretical yield × 100 %
Actual yield = theoretical yield × (% yield/100 %) = 60.25 g × (72 %/100%) = 43 g
