Answer is: 56 gallons of
70% antifreeze and 84 gallons of 95% antifreeze.
ω₁ = 70% ÷ 100% = 0.7; 70% pure antifreeze.
ω₂ = 95% ÷ 100% = 0.95.
ω₃<span> = 85% ÷ 100% = 0.85.
V</span>₁ = ?; volume of 70% antifreeze.
V₂ = ?; volume of 95% antifreeze.<span>
V</span>₃ = V₁ + V₂<span>.
V</span>₃ = 140 gal.
V₁ = 140 gal - V₂<span>.
ω</span>₁ · V₁ + ω₂ ·V₂ = ω₃ · V₃.
0.70 · (140 gal -
V₂) + 0.95 · V₂ = 0.85 · 140 gal.
98 gal - 0.7V₂ + 0.95V₂ = 119 gal.
0.25V₂ = 21 gal.
V₂ = 21 gal ÷ 0.25.
V₂ = 84 gal.
V₁ = 140 gal - 84 gal.
V₁ = 56 gal.
The action is interaction
The mass of dinitrogen tetroxide the number of Nitrogen atoms are 46 x 10^9 gram
Given:
number of nitrogen atoms = 1 x 10^12 atoms
To Find:
mass of dinitrogen tetroxide
Solution:
From the molecular formula of dinitrogen tetraoxide;
molar mass of dinitrogen tetraoxide = 92 g/mol
92 g of dinitrogen tetraoxide contains 2 atoms of nitrogen
x g of dinitrogen tetraoxide will contain 1 * 10^12 atoms of nitrogen
x = 92*1* 10^ 9/2
x = 46 x 10^9 gram
Hence, the number of Nitrogen atoms are
Hence, the number of Nitrogen atoms are 46 x 10^9 gram
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