Calcium carbonate has the formula: CaCO3
From the periodic table:
mass of calcium = 40 grams
mass of carbon = 12 grams
mass of oxygen = 16 grams
Therefore,
molar mass of CaCO3 = 40 + 12 + 3(16) = 100 grams
molar mass of carbonate = 12 + 3(16) = 60 grams
One mole of calcium carbonate contains one mole of carbonate. Therefore, 100 grams of CaCO3 contains 60 grams of CO3.
If the 0.5376 grams of the unknown substance is CaCO3, then the amount of carbonate will be:
amount of carbonate = (0.5376*60) / 100 = 0.32256 grams
Based on the above calculations, the sample is not CaCO3
The answer is c.) Rutherford model
The major visible difference between<span> the two are crystal size, </span>intrusive rocks<span> have a larger crystal/grain texture due to the slow cooling of magma below the earth surface which encourages the growth of larger crystals, while </span>extrusive rocks<span>, because of the rapid cooling at/above the earth's surface does the opposite. Hope I helped</span>
C.<span>The effort force is most often smaller than the force needed without the machine.</span>
Answer:
a) 1.71 × 10⁻³ M
b) 8.00 × 10⁻⁵ M
Explanation:
In order to calculate the solubility (S) of Pb(SCN)₂ we will use an ICE chart. We identify 3 stages (Initial, Change, Equilibrium) and complete each row with the concentration or change in the concentration.
Pb(SCN)₂(s) ⇄ Pb²⁺(aq) + 2 SCN⁻(aq)
I 0 0
C +S +2S
E S 2S
The solubility product (Ksp) is:
Ksp = 2.00 × 10⁻⁵ = [Pb²⁺].[SCN⁻]² = S . (2S)² = 4S³
S = 1.71 × 10⁻³ M
<em>b) Calculate the molar solubility of lead thiocyanate in 0.500 M KSCN.</em>
KSCN is a strong electrolyte that dissociates to give 0.500 M K⁺ and 0.500M SCN⁻.
Pb(SCN)₂(s) ⇄ Pb²⁺(aq) + 2 SCN⁻(aq)
I 0 0.500
C +S +2S
E S 0.500 + 2S
Ksp = 2.00 × 10⁻⁵ = [Pb²⁺].[SCN⁻]² = S . (0.500 + 2S)²
In the term (0.500 + 2S)², 2S is negligible.
Ksp = 2.00 × 10⁻⁵ = S . (0.500)²
S = 8.00 × 10⁻⁵ M
