Answer is: A) 7.84 g.
V(Mg(NO₃)₂) = 151 mL ÷ 1000 mL/L.
V(Mg(NO₃)₂) = 0.151 L; volume of the magnesium nitrate.
c(Mg(NO₃)₂) = 0.352 M; molarity of the solution.
n(Mg(NO₃)₂) = V(Mg(NO₃)₂) · c(Mg(NO₃)₂).
n(Mg(NO₃)₂) ) = 0.151 L · 0.352 mol/L.
n(Mg(NO₃)₂) = 0.0531 mol; amount of the substance.
M(Mg(NO₃)₂) = Ar(Mg) + 2Ar(N) + 6Ar(O) · g/mol.
M(Mg(NO₃)₂) = 24.3 + 2·14 + 6·16 · g/mol.
M(Mg(NO₃)₂) = 148.3 g/mol; molar mass.
m(Mg(NO₃)₂) = n(Mg(NO₃)₂) · M(Mg(NO₃)₂).
m(Mg(NO₃)₂) = 0.0531 mol · 148.3 g/mol.
m(Mg(NO₃)₂) = 7.84; mass of magnesium nitrate.
These could all go either way, hardness and other special properties are what I'm guessing would be the most accurate in determining the kind of material.
luster, cleavage, streak, and color can all be affected by other factors. but I guess cleavage would also be accurate. so I guess hardness special properties and cleavage would be the most reliable.
Answer:
numero de oxidacion: 3+2+2-3
Explanation:
<h2>Hello!</h2>
The answer is:
The new volume is equal to 206.5 L.
<h2>Why?</h2>
To solve this problem, we need to assume that the pressure is constant, and use the Charle's Law equation, so, solving we have:
We are given:
Then, using the Charle's Law equation, we have:
Hence, we have that the new volume is equal to 206.5 L.
Have a nice day!
