The answer is c, relying on renewable energy sources
Stoichiometry:
First, calculate the number of grams for one mole of Ca3 (PO3)4
(3 * (Mass of Ca)) + (4 * (Mass of P + (3 * Mass of Oxygen)))
= (3*40.08) + 4(30.97 + (3*16.00))
=(120.24) + 4(78.97)
=436.12 g / mol Ca3(PO3)4
This means there are 436.12 g per 1 mole of Ca(PO3)4. Since there are 4.50 moles of Calcium Phosphate, mulitply the molar mass of Ca(PO3)4 by 4.50 and you should get 1962.54 g. Since there are 3 sigfigs, the final answer is 1960 g.
on a side note: I put in all my work in case 1. your periodic table if different, 2. my work is wrong, 3. you put in the question wrong because I feel that the actual compound would be Ca3(PO4)3 instead of Ca3(PO3)4 (if this is the case, the answer should be 1820 g).
The correct answer for this question is
20.2 g silver placed in 21.6 mL of water and 12.0 g silver placed in 21.6 mL of water.
15.2 g copper placed in 21.6 mL of water and 50.0 g copper placed in 23.4 mL of water.
11.2 g gold placed in 21.6 mL of water and 14.9 g gold placed in 23.4 mL of water.
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Answer:
1.7 Osm/L
Explanation:
Osmotic concentration is the number of osmoles (Osm) of solute per litre of solution .
An osmole is the number of moles of solute that contribute to the osmotic pressure of a solution.
1 Osm = i × n, where
i = the van't Hoff i factor and
n = the number of moles
The van't Hoff i factor is the number of solute particles obtained from 1 mol of solute.
For example,
MgSO₄(aq) ⟶ Mg²⁺(aq) + SO₄²⁻(aq)
1 mol of MgSO₄ produces 2 mol of ions in solution, so i = 2.
The product formed when benzene reacts with isobutyl chloride in the presence of AlCl3 is tertiary butyl benzene.
THE CHEMICAL REACTION AS
step 1 : CH₃ - CH - CH₂Cl + AlCl₃ → CH₃ - CH₂ -CH₂ - CH₂⁺ + AlCl⁻
step 2: CH₃ - CH₂ - CH₂ - CH₂⁺ ---H SHIFT--→ (CH₃)₃C⁺
PRODUCT
C₆H₆ + (CH₃)₃ C⁺ → C₆H₆ ( CH₃)₃ C⁺
Hence the product formed is tertiary butyl benzene.
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