<h3>
Answer:</h3>
3CaCl₂ + 2Na₃PO₄→ Ca₃(PO₄)₂ + 6NaCl
<h3>
Explanation:</h3>
We are given the Equation;
CaCl₂ + Na₃PO₄→ Ca₃(PO₄)₂ + NaCl
Assuming the question requires us to balance the equation;
- A balanced chemical equation is one that has equal number of atoms of each element on both sides of the equation.
- Balancing chemical equations ensures that they obey the law of conservation of mass in chemical equations.
- According to the law of conservation of mass in chemical equation, the mass of the reactants should always be equal to the mass of the products.
- Balancing chemical equations involves putting appropriate coefficients on the reactants and products.
In this case;
- To balance the equation we are going to put the coefficients 3, 2, 1, and 6.
- Therefore; the balanced equation will be;
3CaCl₂ + 2Na₃PO₄→ Ca₃(PO₄)₂ + 6NaCl
Answer:
Percent yield = 89.1%
Explanation:
Based on the equation:
Cl₂ + 2KI → 2KCl + I₂
<em>1 mole of Cl₂ reacts with 2 moles of KI to produce to moles of KCl</em>
<em />
To solve this quesiton we must find the moles of each reactant in order to find the limiting reactant. With the limiting reactant we can find the moles of KCl and the mass:
<em>Moles Cl₂:</em>
8x10²⁵ molecules * (1mol / 6.022x10²³ molecules) = 133 moles
<em>Moles KI -Molar mass: 166.0028g/mol-</em>
25g * (1mol / 166.0028g) = 0.15 moles
Here, clarely, the KI is the limiting reactant
As 2 moles of KI produce 2 moles of KCl, the moles of KCl produced are 0.15 moles. The theoretical mass is:
0.15 moles * (74.5513g / mol) =
11.2g KCl
Percent yield is: Actual yield (10.0g) / Theoretical yield (11.2g) * 100
<h3>Percent yield = 89.1%</h3>
Answer:
semipermeability
Explanation:
partially but not freely or wholly permeable specifically : permeable to some usually small molecules but not to other usually larger particles a semipermeable membrane.
Answer:
697 g
Explanation:
Ethanol (C₂H₅OH) and butanoic acid (C₃H₇COOH) react to form ethyl butanoate (C₃H₇COOC₂H₅) and water (H₂O).
C₂H₅OH + C₃H₇COOH → C₃H₇COOC₂H₅ + H₂O
The molar ratio of C₂H₅OH to C₃H₇COOC₂H₅ is 1:1. The moles of C₃H₇COOC₂H₅ produced from 6.00 moles of C₂H₅OH are:
6.00 mol C₂H₅OH × (1 mol C₃H₇COOC₂H₅/1 mol C₂H₅OH) = 6.00 mol C₃H₇COOC₂H₅
The molar mass of C₃H₇COOC₂H₅ is 116.16 g/mol. The mass corresponding to 6.00 mol is:
6.00 mol × (116.16 g/mol) = 697 g
