Answer:
2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.
Explanation:
Mass of fat = 2.4 kg = 2.4 × 1000 g = 2400 g
1 kg = 1000 g
Molar mass of fat = M
M = 57 × 12 g/mol + 110 × 1 g/mol+ 6 × 16 g/mol = 890 g/mol[/tex]
Moles of fat = 
According to reaction , 2 moles of fat gives 110 moles of water. Then 2.6966 moles of fat will give ;
of water
Mass of 148.31 moles of water ;
148.31 mol × 18 g/mol = 2,669.58 g
2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.
Planets, black wholes gravity
Answer:
[Ag⁺] = 5.0x10⁻¹⁴M
Explanation:
The product solubility constant, Ksp, of the insoluble salts PbI₂ and AgI is defined as follows:
Ksp(PbI₂) = [Pb²⁺] [I⁻]² = 1.4x10⁻⁸
Ksp(AgI) = [Ag⁺] [I⁻] = 8.3x10⁻¹⁷
The PbI₂ <em>just begin to precipitate when the product [Pb²⁺] [I⁻]² = 1.4x10⁻⁸</em>
<em />
As the initial [Pb²⁺] = 0.0050M:
[Pb²⁺] [I⁻]² = 1.4x10⁻⁸
[0.0050] [I⁻]² = 1.4x10⁻⁸
[I⁻]² = 1.4x10⁻⁸ / 0.0050
[I⁻]² = 2.8x10⁻⁶
<h3>[I⁻] = 1.67x10⁻³</h3><h3 />
So, as the [I⁻] concentration is also in the expression of Ksp of AgI and you know concentration in solution of I⁻ = 1.67x10⁻³M:
[Ag⁺] [I⁻] = 8.3x10⁻¹⁷
[Ag⁺] [1.67x10⁻³] = 8.3x10⁻¹⁷
<h3>[Ag⁺] = 5.0x10⁻¹⁴M</h3>
B
. They are large and occur at shallow depths near the where the plates diverge.
Answer:
2.33 mol C
Explanation:
Step 1: Write the balanced generic chemical equation
3 A ⟶ C + 4 D
Step 2: Establish the appropriate molar ratio
According to the balanced equation, the molar ratio of A to C is 3:1.
Step 3: Calculate the number of moles of C produced from 7 moles of A
We will use the previously established molar ratio.
7 mol A × 1 mol C/3 mol A = 2.33 mol C
