218.4 grams of CaO is produced using 3.9 moles CaCO₃.
<h3>How we calculate weight of any substance from moles?</h3>
Moles of any substance will be defined as:
n = W / M
Given chemical reaction is:
CaCO₃ → CaO + CO₂
From the above equation it is clear that according to the concept of stoichiometry 1 mole of CaCO₃ is producing 1 mole of CaO. By using above formula, we calculate grams as follow:
W = n × M, where
n = no. of moles of CaO = 3.9 moles
M = molar mass of CaO = 56 g/mole
W = 3.9 × 56 = 218.4 g
Hence, 218.4 grams of Cao is produced.
<h3>How much grams do Cao's molecular weight equal?</h3>
Molecular weight of CaO. CaO has a molar mass of 56.0774 g/mol. Calcium Oxide is another name for this substance. Convert moles of CaO to grams or grams of CaO to moles. Calculation of the molecular weight: 40.078 + 15.9994 ›› Composition by percentage and element
<h3>How much calcium is required to create one mole of oxygen?</h3>
In order to create one mole of calcium oxide, the reaction between one mole of calcium (40.1 g) and half a mole of oxygen (16 g) is stoichiometric (56.1 g). This means that only 4.01 grams of calcium metal and 1.6 grams of oxygen can combine to generate 5.61 grams of calcium oxide.
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Answer:
The molar mass of the gas is 44.19 g/mol
Explanation:
Amount of sample of gas = m = 13.5 g
Volume occupied by the gas = V = 5.10 L
Pressure of the gas = P = 149.83 KPa
1 KPa = 0.00986 atm
P = 
Assuming M g/mol to be the molar mass of the gas
Assuming the gas is behaving as an ideal gas

The molar mass of gas is 44.19 g/mol
Catalyst can speed up chemical reaction without being used up or involved in thr reaction..
Answer:
Explanation:
Given parameters;
pH = 8.74
pH = 11.38
pH = 2.81
Unknown:
concentration of hydrogen ion and hydroxyl ion for each solution = ?
Solution
The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.
It is graduated from 1 to 14
pH = -log[H₃O⁺]
pOH = -log[OH⁻]
pH + pOH = 14
Now let us solve;
pH = 8.74
since pH = -log[H₃O⁺]
8.74 = -log[H₃O⁺]
[H₃O⁺] = 10⁻
[H₃O⁺] = 1.82 x 10⁻⁹mol dm³
pH + pOH = 14
pOH = 14 - 8.74
pOH = 5.26
pOH = -log[OH⁻]
5.26 = -log[OH⁻]
[OH⁻] = 10
[OH⁻] = 5.5 x 10⁻⁶mol dm³
2. pH = 11.38
since pH = -log[H₃O⁺]
11.38 = -log[H₃O⁺]
[H₃O⁺] = 10⁻
[H₃O⁺] = 4.17 x 10⁻¹² mol dm³
pH + pOH = 14
pOH = 14 - 11.38
pOH = 2.62
pOH = -log[OH⁻]
2.62 = -log[OH⁻]
[OH⁻] = 10
[OH⁻] =2.4 x 10⁻³mol dm³
3. pH = 2.81
since pH = -log[H₃O⁺]
2.81 = -log[H₃O⁺]
[H₃O⁺] = 10⁻
[H₃O⁺] = 1.55 x 10⁻³ mol dm³
pH + pOH = 14
pOH = 14 - 2.81
pOH = 11.19
pOH = -log[OH⁻]
11.19 = -log[OH⁻]
[OH⁻] = 10
[OH⁻] =6.46 x 10⁻¹²mol dm³
Answer:
See explanation
Explanation:
A balanced chemical reaction equation has the same number of atoms of each element on both sides of the reaction equation.
Hence, for the reaction between KOH and H2SO4, the balanced chemical reaction equation is;
H2SO4(aq) + 2KOH(aq) ---------> K2SO4(aq) + 2H2O(l)
Complete ionic equation;
2H^+(aq) + SO4^2-(aq) + 2K^+(aq) +2OH^-(aq) -------> SO4^2-(aq) + 2K^+(aq) + 2H2O(l)
Net ionic equation;
2H^+(aq) + 2OH^-(aq) -------> 2H2O(l)