Answer:
4777.09grams
Explanation:
To find the mass of oxygen, we first need to find the mass of potassium sulfate (K2SO4).
Since there are 4.50 x 10^25 formula units of potassium sulfate, we can find the number of moles in K2SO4 by dividing by Avagadros number (6.02 × 10^23 units). That is;
number of moles of K2SO4 (n) = 4.50 x 10^25 ÷ 6.02 × 10^23
= 0.747 × 10^ (25-23)
= 0.747 × 10^2
= 7.47 × 10^1 moles
Mass in grams of K2SO4 can be calculated thus: molar mass of K2SO4 × moles
= 174.252 g/mol × 7.47 × 10^1 moles
= 13016.62grams.
Since mass of oxygen in 1 mol of K2SO4 = O4 = 63.996 g/mol
We find the percentage by mass of oxygen in K2SO4 as follows:
= 63.996/174.252 × 100
= 0.367 × 100
= 36.7% by mass of oxygen.
This means that in 1gram of K2SO4, there are 0.367gram of Oxygen. Hence, in 13016.62grams (4.50 x 10^25 units) of K2SO4, there will be;
0.367 × 13016.62
= 4777.09grams of oxygen.
Answer:
The enthalpy change during the reaction is -199. kJ/mol.
Explanation:
Mass of solution = m
Volume of solution = 100.0 mL
Density of solution = d = 1.00 g/mL
First we have to calculate the heat gained by the solution in coffee-cup calorimeter.
where,
m = mass of solution = 100 g
q = heat gained = ?
c = specific heat =
= final temperature =
= initial temperature =
Now put all the given values in the above formula, we get:
Now we have to calculate the enthalpy change during the reaction.
where,
= enthalpy change = ?
q = heat gained = 2.242 kJ
n = number of moles fructose =
Therefore, the enthalpy change during the reaction is -199. kJ/mol.
Answer is:
concentration of ammonium ions are 7,14·10⁻¹⁴ M.<span>
Chemical reaction: 2NH</span>₃(l) → NH₄⁺(am) + NH₂⁻(am).<span>
Kam = 5,1·10</span>⁻²⁷.<span>
[NH</span>₄⁺] ·
[NH₂⁻] = x; equilibrium concentration of cations and anions.<span>
Kam = [NH</span>₄⁺] · [NH₂⁻].<span>
Kam = x².
x = [NH</span>₄⁺]
= √5,1·10⁻²⁷.<span>
[NH</span>₄⁺] =
7,14·10⁻¹⁴<span> M.</span>
Answer:
0.01504 mol/dm3
Explanation:
Calcium Hydroxide - Ca(OH)2
Hydrochloric acid - HCl
The reaction between both is given as;
2HCl + Ca(OH)2 --> CaCl2 + 2H2O
2 : 1
The relationship between the concentration and volume is given as;
CaVa / CbVb = na / nb
where a = acid (HCl) and b = base Ca(OH)2
Inserting the values;
0.04 * 18.8 / (Cb * 25) = 2 / 1
0.752 / 25Cb = 2
Cb = 0.752 / 50 = 0.01504
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