You must add 45 mL of the 80 % alcohol to the 30 % alcohol to get a 35 % solution.
You can use a modified dilution formula to calculate the volume of 80 % alcohol
V1×C1 + V2×C2 = V3×C3
Let the volume of 80 % mixture 1 = <em>x</em> mL. Then the volume of the final 35 % mixture 3 = (405 + <em>x</em> ) mL
(<em>x</em> mL×80 % alc) + (405 mL×30 % alc) = (405 + <em>x</em>)mL × 35 % alc
80x + 12 150 = 14 175 + 35 x
45x = 2025
x = 2025/45 = 45
The balanced chemical equation is :
5P₄ + 36OH → 12HPO₃⁻² (aq) + 8PH₃ (acidic)
Here the oxidation number of P changed from 0 to -3 in PH₃ and increases from 0 to +3 in HPO₃⁻². When P₄ changes to PH₃ reduction reaction is taking place as there is addition of hydrogen and when P₄ changes to HPO₃⁻² oxidation takes place as there is addition of oxygen.
Thus clearly both reduction and oxidation are taking place.
Thus, we can infer that here P₄ is both oxidizing as well as reducing agent.
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It depends if you are talking about the ones that go up or down
or left and right
the ones that go up and down are called groups
the onesthat go left and right are periods
160 g of SO3 are needed to make 400 g of 49% H2SO4.
<h3>How many grams of SO3 are required to prepare 400 g of 49% H2SO4?</h3>
The equation of the reaction for the formation of H2SO4 from SO3 is given below as follows:

1 mole of SO3 produces 1 mole of H2SO4
Molar mass of SO3 = 80 g/mol
Molar mass of H2SO4 = 98 g/mol
80 g of SO3 are required to produce 98 og 100%H2SO4
mass of SO3 required to produce 400 g of 100 %H2SO4 = 80/98 × 400 = 326.5 g of SO3
Mass of SO3 required to produce 49% of 400 g H2SO4 = 326.5 × 49% = 160 g
Therefore, 160 g of SO3 are needed to make 400 g of 49% H2SO4.
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