The fertilizer contains 7.90 % P by mass.
<em>Assume</em> you have 100 g of fertilizer.
<em>Mass of P₄O₁₀</em> = 100 g fertilizer × 18.1 g P₄O₁₀/100 g fertilizer = 18.1 g P₄O₁₀
<em>M_r of P₄O₁₀</em> = 4 × 30.97 + 10× 16.00 = 123.88 + 160.00 = 283.88
<em>Mass of P</em> = 18.1 g P₄O₁₀ × (123.88 g P/283.88 g P₄O₁₀) = 7.90 g P
<em>Mass % P</em> = mass/total mass × 100 % = 7.90 g /100 g × 100 % = 7.90 %
Answer:
true
Explanation:
cffffyifuducxirrtkdyfufiuruhoy
Answer:
E. The ratio of [HNO2] / [NO2-] will increase
D. The pH will decrease.
Explanation:
Nitrous acid ( HNO₂ ) is a weak acid and NaNO₂ is its salt . The mixture makes a buffer solution .
pH = pka + log [ salt] / [ Acid ]
= 3.4 + log .32 / .42
= 3.4 - .118
= 3.282 .
Now .16 moles of nitric acid is added which will react with salt to form acid
HNO₃ + NaNO₂ = HNO₂ + NaNO₃
concentration of nitrous acid will be increased and concentration of sodium nitrite ( salt will decrease )
concentration of nitrous acid = .42 + .16 = .58 M
concentration of salt = .32 - .16 = .16 M
ratio of [HNO₂ ] / NO₂⁻]
= .42 / .32 = 1.3125
ratio of [HNO₂ ] / NO₂⁻] after reaction
= .42 + .16 / .32 - .16
= 58 / 16
= 3.625 .
ratio will increase.
Option E is the answer .
pH after reaction
= 3.4 + log .16 / .58
= 2.84
pH will decrease.
<u>Answer:</u>
<u>Explanation:</u>
Frozen water is known to have more volume, but less density. When water slowly gets unfrozen, it'd volume decreases and its density increases. Hence, when the water gets unfrozen, it's volume decreases.
