Answer:
0.049 mol/L.s
Explanation:
The decomposition of hydrogen peroxide is:

![Rate = -\dfrac{\Delta [H_2O_2]}{\Delta t}= \dfrac{\Delta [H_2O_2]}{\Delta t}= \dfrac{ 2 \Delta [H_2O_2]}{\Delta t}](https://tex.z-dn.net/?f=Rate%20%3D%20-%5Cdfrac%7B%5CDelta%20%5BH_2O_2%5D%7D%7B%5CDelta%20t%7D%3D%20%5Cdfrac%7B%5CDelta%20%5BH_2O_2%5D%7D%7B%5CDelta%20t%7D%3D%20%5Cdfrac%7B%202%20%20%5CDelta%20%5BH_2O_2%5D%7D%7B%5CDelta%20t%7D)
The rate of decomposition reaction = the rate of formation of
= 0.098 mol/L.s
∴
Rate of formation of


= 0.049 mol/L.s
Answer:
Temperature = 44.02°C
Explanation:
Insulated container indicates no heat loss to the surroundings.
The specific heat capacity of a substance is a physical property of matter. It is defined as the amount of heat that is to be supplied to a unit mass of the material to produce a unit change in its temperature.
The SI unit of specific heat is joule per kelvin and kilogram, J/(K kg).
Now,
Specific heat for water is 4.1813 Jg⁻¹K⁻¹.
Latent heat of vaporization of water is 2257 Jg⁻¹.
Energy lost by steam in it's process of conversion to water, is the energy acquired by water resulting in an increase in it's temperature.

Q= Heat transferred
m= mass of the substance
T= temperature
Also,

L= Latent heat of fusion/ vaporization ( during phase change)
Now applying the above equations to the problem:


Temperature = 44.02°C
Answer:
1. 0.74mol
2. 0.42mol
3. 2.125mol
4. 0.301mol
5. 4.52 × 10^23 particles
Explanation:
Number of moles (n) in a substance can be found using the formula:
mole (n) = mass/molar mass
Using this formula, the following moles are calculated:
1. Molar of Na = 23g/mol
mole = 17/23
mole = 0.74mol
2. Molar mass of Na2SO4 = 23(2) + 32 + 16(4)
= 46 + 32 + 64
= 142g/mol
Mole = 60/142
mole = 0.42mol
3. Molar mass of CO2 = 12 + 16(2)
= 12 + 32
= 44g/mol
mole = 93.5/44
mole = 2.125mol
4. Molar mass of sodium nitrate (NaNO3) = 23 + 14 + 16(3)
= 23 + 14 + 48
= 85g/mol
mole = 25.6/85
mole = 0.301mol
5. Number of particles in one mole of a substance is 6.022 × 10^23 particles. Hence, in 0.75mol of calcium hydroxide (Ca(OH)2, there will be;
0.75mol × 6.02 × 10^23
= 4.515 × 10^23
= 4.52 × 10^23 particles
Answer:
0.30 mol/L
Explanation:
Mass = 108 g
Molar mass of glucose = 180.156 g/mol
The formula for the calculation of moles is shown below:
Thus,

Given Volume = 2 L
<u>Molarity = 0.3 mol/L</u>