Answer: The weight/weight % or percent by mass of the solute is 5.41 %.
Explanation:
Mass of the sodium sulfate,w = 9.74 g
Volume of the water = 165 mL
Density of the water = 1 g/mL
Mass of the water =
Mass of the solution, W:
Mass of solute + Mass of solvent =9.47 g + 165 g=174.47 g
The weight/weight % or percent by mass of the solute is 5.41 %.
The balanced equation given is:
4NH3 + 3O2 .....> 2N2 + 6H2O
From this equation, we can note that 4 moles of NH3 are required to produce 2 moles of N2.
Therefore, the mole ratio of NH3 to N2 is 4:2 which can be simplified into 2:1
Answer:
There are 3 steps of this problem.
Explanation:
Step 1.
Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.
Step 2.
Enthalpy of saturated liquid Haq = 781.124 J/g
Enthalpy of saturated vapour Hvap = 2779.7 J/g
Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g
Step 3.
In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy
So, H1=H2
H2= (1-x)Haq+XHvap.........1
Putting the values in 1
2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}
= 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)
1904.976 (J/g) = x1998.576 (J/g)
x = 1904.976 (J/g)/1998.576 (J/g)
x = 0.953
So, the quality of the wet steam is 0.953
Answer:
The answer is 465.6 mg of MgI₂ to be added.
Explanation:
We find the mole of ion I⁻ in the final solution
C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol
But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.
So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.
Hence, the weight of MgI₂ must be added is
Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg
Flame colors are produced from the movement of the electrons in the metal ions present in the compounds. When you heat it, the electrons gain energy and can jump into any of the empty orbitals at higher levels Each of these jumps involves a specific amount of energy being released as light energy, and each corresponds to a particular color. As a result of all these jumps, a spectrum of colored lines will be produced. The color you see will be a combination of all these individual colors.
