1.9264*10^23 is the answer
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.
Learn more about moles:
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Answer:
<h2>hail</h2>
Explanation:
<h3><u>Large updrafts of air can throw rain droplets high up into the tops of the cloud.</u></h3>
<em>Answer:</em>
The new weight-loss supplement was less effective on females in the study. That is because only 65% of them lost weight after two weeks, whereas 74% of the males lost weight after two weeks.
<em>Explanation:</em>
37 out of 50 males lost weight after two weeks, so 74% of males lost weight ( (37 · 2)/(50 · 2) ). 13 out of 20 females lost weight after two weeks, so 65% of females lost weight ( (13 · 5)/(20 · 5) ). This means the supplement was less effective on females because males had a higher percentage of losing weight.
Hope this helps !! :D
When PH = ㏒(H^+)
[H+] = 10^-3.67
[H+] = 2.14x 10^-4
so according to the reaction equation:
HF ↔ H^+ + F^-
at equilibrium X-(2.14x10^-4) (2.14x10^-4) (2.14x10^-4)
by substitution in Ka formula:
Ka = [H+][F]-/[HF]
6.8x10^-4 = (2.14x10^-4)*(2.14x10^-4) /(X-2.4x10^-4)
X-2.4x10^-4 = (4.58x10^-8)* (6.8x10^-4)
∴X = 2.4x10^-4
∴[HF] = X- (2.14x10^-4)= (2.4x10^-4)-(2.14x10^-4) = 2.6 X 10^-5