25 sugar because the mass should be equal on both sides
Answer:
=759.95 grams.
Explanation:
The molar mass of chromium is 51.9961 g/mol
Therefore the number of moles of chromium in 156 grams is:
Number of moles =mass/RAM
=156g/51.9961g/mol
=3 moles.
From the equation provided, 3 moles of chromium metal produce 2 moles of Chromium oxide.
Therefore 3 moles of chromium produce:
(3×2)/4 moles =1.5 moles of chromium oxide.
I mole of chromium oxide has a mass of 151.99 g
Thus 1.5 moles= 1.5mole ×151.99 g/mol
=759.95 grams.
<u>Answer:</u> The mass percent of hydrogen in methyl acetate is 8 %
<u>Explanation:</u>
The given chemical formula of methyl acetate is 
To calculate the mass percentage of hydrogen in methyl acetate, we use the equation:

Mass of hydrogen = (6 × 1) = 6 g
Mass of methyl acetate = [(3 × 12) + (6 × 1) + (2 × 16)] = 74 g
Putting values in above equation, we get:

Hence, the mass percent of hydrogen in methyl acetate is 8 %
Since gold and silver are the least reactive metals, they do not react with water. The surface of metallic lead is covered by a thin layer of lead oxide,. As a result, it does not react with water in normal circumstances.
Explanation:
<h3>Hope it helps you!! </h3>
1) Chemical reaction:
2(CH₃COO)₃Fe + 3MgCrO₄ → Fe₂(CrO₄)₃ + 3(CH₃COO)₂Mg.
m((CH₃COO)₃Fe) = 15,0 g.
m(MgCrO₄) = 10,0 g.
n((CH₃COO)₃Fe) = m((CH₃COO)₃Fe) ÷ M((CH₃COO)₃Fe).
n((CH₃COO)₃Fe) = 15 g ÷ 233 g/mol.
n((CH₃COO)₃Fe) = 0,064 mol.
n(MgCrO₄) = m(MgCrO₄) ÷ M(MgCrO₄).
n(MgCrO₄) = 10 g ÷ 140,3 g/mol.
n(MgCrO₄) = 0,071 mol; limiting reagens.
From chemical reaction: n(MgCrO₄) : n((CH₃COO)₂Mg) = 3 : 3.
n((CH₃COO)₂Mg) = 0,071 mol.
m((CH₃COO)₂Mg) = 0,071 mol · 142,4 g/mol.
n((CH₃COO)₂Mg) = 10,11 g.
2) Chemical reaction:
2(CH₃COO)₃Fe + 3MgSO₄ → Fe₂(SO₄)₃ + 3(CH₃COO)₂Mg.
m((CH₃COO)₃Fe) = 15,0 g.
m(MgSO₄) = 15,0 g.
n((CH₃COO)₃Fe) = m((CH₃COO)₃Fe) ÷ M((CH₃COO)₃Fe).
n((CH₃COO)₃Fe) = 15 g ÷ 233 g/mol.
n((CH₃COO)₃Fe) = 0,064 mol; limiting ragens.
n(MgSO₄) = m(MgSO₄) ÷ M(MgSO₄).
n(MgSO₄) = 15 g ÷ 120,36 g/mol.
n(MgSO₄) = 0,125 mol; limiting reagens.
From chemical reaction: n(CH₃COO)₃Fe) : n((CH₃COO)₂Mg) = 2 : 3.
n((CH₃COO)₂Mg) = 0,064 mol · 3/2.
n((CH₃COO)₂Mg) = 0,096 mol.
m((CH₃COO)₂Mg) = 0,096 mol · 142,4 g/mol.
m((CH₃COO)₂Mg) = 13,7 g.