Answer: The empirical formula is
Explanation:
Mass of Mg= 46.6 g
Mass of S= 61.4 g
Mass of O = 92.0 g
Step 1 : convert given masses into moles.
Moles of Mg =
Moles of S =
Moles of O =
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Mg =
For S=
For O =
The ratio of Mg: S: O = 1: 1: 3
Hence the empirical formula is
Is that the question orr......?
Answer:
True.
Explanation:
this is because undisturbed sedimentary rocks forms in layers.
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 concentration of nitrate ion is 0.033 M.
Explanation:
1 mole of = 3 mole
Let = 3y
then according to above reaction,
= y
= 0.10 M (given)
= ?
3y = 0.10 M
y = 0.033 M
So ,concentration of = y = 0.033 M