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
Mass =?
moles of N2 = 4.25 x 103 mol
molar mass of N2 = (14)x2 = 28
Answer:
a. 2 HgO(s) ⇒ 2 Hg(l) + O₂(g)
b. 0.957 g
Explanation:
Step 1: Write the balanced equation
2 HgO(s) ⇒ 2 Hg(l) + O₂(g)
Step 2: Convert 130.0 °C to Kelvin
We will use the following expression.
K = °C + 273.15
K = 130.0°C + 273.15
K = 403.2 K
Step 3: Calculate the moles of O₂
We will use the ideal gas equation.
P × V = n × R × T
n = P × V/R × T
n = 1 atm × 0.0730 L/0.0821 atm.L/mol.K × 403.2 K
n = 2.21 × 10⁻³ mol
Step 4: Calculate the moles of HgO that produced 2.21 × 10⁻³ moles of O₂
The molar ratio of HgO to O₂ is 2:1. The moles of HgO required are 2/1 × 2.21 × 10⁻³ mol = 4.42 × 10⁻³ mol.
Step 5: Calculate the mass corresponding to 4.42 × 10⁻³ moles of HgO
The molar mass of HgO is 216.59 g/mol.
4.42 × 10⁻³ mol × 216.59 g/mol = 0.957 g
Answer:
Faraday's constant will be smaller than it is supposed to be.
Explanation:
If the copper anode was not completely dry when its mass was measured, mass of the copper must be heavier than it should have been. Hence, the calculated Faraday’s constant would be smaller than it is supposed to be since when calculating Faraday’s Constant, the charge transferred is divided by the moles of electrons.
Answer is: the volume of an irregular object is 4,00 ml.
<span>Volume is the amount of space the object occupies and can be finded immersing it in water in a container with volume markings and than see how much the level of the container changes (goes up).
</span>V(irregular object) = V(final volume) - V(initial volume).
V(irregular object) = 7,50 ml - 3,50 ml.
V(irregular object) = 4,00 ml.
