1s^2 = filled
2s^2 = filled
2p^6 = filled
3s^1 = partially filled (need 2)
3 orbitals are completely filled.
Answer:
c
Explanation:
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CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial Y - -
Change -X +X +X
Equilibrium Y-X X X
Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²
Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
X = 5.79 x 10⁻⁵ M
Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
= 5.79 x 10⁻⁵ mol/L
Molar mass of CaCO₃ = 100 g mol⁻¹
Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
= 5.79 x 10⁻³ g/L
Answer:
Explanation:
From the statement of the problem,
B₂S₃
+ H₂O
→ H₃BO₃
+ H₂S
B₂S₃ + H₂O → H₃BO₃ + H₂S
We that the above expression does not conform with the law of conservation of mass:
To obey the law, we need to derive a balanced reaction equation:
Let us use the mathematical method to obtain a balanced equation.
let the balanced equation be:
aB₂S₃ + bH₂O → cH₃BO₃ + dH₂S
where a, b, c and d will make the equation balanced.
Conservating B: 2a = c
S: 3a = d
H: 2b = 3c + 2d
O: b = 3c
if a = 1,
c = 2,
b = 6,
2d = 2(6) - 3(2) = 6, d = 3
Now we can input this into our equation:
B₂S₃ + 6H₂O → 2H₃BO₃ + 3H₂S
B₂S₃
+ 6H₂O
→ 2H₃BO₃
+ 3H₂S
Answer:
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Explanation: