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Answer:
A solution is made by dissolving 4.87 g of potassium nitrate in water to a final volume of 86.4 mL solution. The weight/weight % or percent by mass of the solute is :
<u>2.67%</u>
Explanation:
Note : Look at the density of potassium nitrate in water if given in the question.
<u><em>You are calculating </em></u><u><em>weight /Volume</em></u><u><em> not weight/weight % or percent by mass of the solute</em></u>
Here the <u>weight/weight % or percent by mass</u> of the solute is asked : So first convert the<u> VOLUME OF SOLUTION into MASS</u>
Density of potassium nitrate in water KNO3 = 2.11 g/mL

Density = 2.11 g/mL
Volume of solution = 86.4 mL



Mass of Solute = 4.87 g
Mass of Solution = 183.2 g
w/w% of the solute =


w/w%=2.67%
= 9.1 × 10^6
(scientific notation)
= 9.1e6
(scientific e notation)
= 9.1 × 10^6
(engineering notation)
(million; prefix mega- (M))
= 9100000
<span>(real number)</span>
Answer:
P₂ ≅ 100 atm (1 sig. fig. based on the given value of P₁ = 90 atm)
Explanation:
Given:
P₁ = 90 atm P₂ = ?
V₁ = 18 Liters(L) L₂ = 12 Liters(L)
=> decrease volume => increase pressure
=> volume ratio that will increase 90 atm is (18L/12L)
T₁ = 272 Kelvin(K) T₂ = 274 Kelvin(K)
=> increase temperature => increase pressure
=> temperature ratio that will increase 90 atm is (274K/272K)
n₁ = moles = constant n₂ = n₁ = constant
P₂ = 90 atm x (18L/12L) x (274K/272K) = 135.9926471 atm (calculator)
By rule of sig. figs., the final answer should be rounded to an accuracy equal to the 'measured' data value having the least number of sig. figs. This means P₂ ≅ 100 atm based on the given value of P₁ = 90 atm.