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:
Explanation:
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In this case, according to the Avogadro's number, it is possible to compute the atoms of Kr in 2.00 moles as shown below:
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Answer:
CuBr₂(aq) + Pb(CH₃COO)₂(aq) → Cu(CH₃COO)₂(aq) + PbBr₂ (s)↓
Explanation:
We identify the reactants:
CuBr₂ and Pb(CH₃COO)₂
The products will be: Cu(CH₃COO)₂ and PbBr₂
You may know these information:
Salts from acetate are soluble.
Bromide can make solid salts with these cations: Ag⁺, Pb²⁺, Hg₂²⁺, Cu⁺
PbBr₂ is formed, so this will be our precipitate
The equation is:
CuBr₂(aq) + Pb(CH₃COO)₂(aq) → Cu(CH₃COO)₂(aq) + PbBr₂ (s)↓
