0,35 kmol/m³ = 0,35 mol/dm³ = 0,35 mol/L
175 mL = 0,175 L
*-*-*-*-*-*-*-*-*-*-*-*
C = n/V
n = 0,35×0,175
n = 0,06125 mol
mCa(NO₃)₂: 40+(14×2)+(16×6) = 164 g/mol
1 mol --------- 164g
0,06125 ---- X
X = 10,045g
To prepare 175 mL of 0,35M solution, add 10,045g of calcium nitrate and add water to a volume of 175ml.
Answer:
0.0905 M
Explanation:
Let's consider the neutralization reaction between H2SO4 and KOH.
H₂SO₄ + 2 KOH → K₂SO₄ + 2 H₂O
22.87 mL of 0.158 M KOH react. The reacting moles of KOH are:
0.02287 L × 0.158 mol/L = 3.61 × 10⁻³ mol
The molar ratio of H₂SO₄ to KOH is 1:2. The reacting moles of H₂SO₄ are 1/2 × 3.61 × 10⁻³ mol = 1.81 × 10⁻³ mol
1.81 × 10⁻³ moles of H₂SO₄ are in 20.0 mL. The molarity of H₂SO₄ is:
M = 1.81 × 10⁻³ mol / 0.0200 L = 0.0905 M
Answer:
All of them!
Explanation:
Since Mg, Li, Ca, and Cs are all in groups 1 and 2 of the periodic table, they are alkali/alkaline earth metals and will all lose electrons during ionic bonding.
D) because both reactions are occurring at the same rate. They are not equal but their concentrations are constant.
Answer:
A
Explanation:
because it loses 2 electrons
