Answer:
im pretty sure its the second one
Answer:
5.0 moles of water per one mole of anhydrate
Explanation:
To solve this question we must find the moles of the anhydrate. The difference in mass between the dry and the anhydrate gives the mass of water. Thus, we can find the moles of water and the moles of water per mole of anhydrate:
<em>Moles Anhydrate:</em>
7.58g * (1mol / 84.32g) = 0.0899 moles XCO3
<em>Moles water:</em>
15.67g - 7.58g = 8.09g * (1mol / 18.01g) = 0.449 moles H2O
Moles of water per mole of anhydrate:
0.449 moles H2O / 0.0899 moles XCO3 =
5.0 moles of water per one mole of anhydrate
Answer:
The specific heat of gold is 0.129 J/g°C
Explanation:
Step 1: Data given
Mass of gold = 15.3 grams
Heat absorbed = 87.2 J
Initial temperature = 35.0 °C
Final temperature = 79.2 °C
Step 2:
Q = m*c*ΔT
⇒ Q =the heat absorbed = 87.2 J
⇒ m = the mass of gold = 15.3 grams
⇒ c = the specific heat of gold = TO BE DETERMINED
⇒ ΔT = The change in temperature = T2 - T1 = 79.2 - 35.0 = 44.2 °C
87.2 J = 15.3g * c * 44.2°C
c = 87.2 / (15.3 * 44.2)
c = 0.129 J/g°C
The specific heat of gold is 0.129 J/g°C
Answer
is: The molar solubility of calcium phosphate is 108s⁵ = Ksp.
<span>
Balanced chemical reaction: Ca</span>₃(PO₄)₂(s) → 3Ca²⁺(aq) + 2PO₄³⁻(aq).<span>
[Ca²</span>⁺] =
3s(Ca₃(PO₄)₂) =
3s.<span>
[PO</span>₄³⁻] = 2s.<span>
Ksp = [Ca²</span>⁺]³ · [PO₄³⁻]².<span>
Ksp = (3s)³ · (2s)².
Ksp = 108s</span>⁵.
s = ⁵√(Ksp ÷ 108).
