A flask with a volume of 125.0 mL contains air with a density of 1.298 g/L. what is the mass of the air contained in the flask<span>The given are: </span>
<span><span>1. </span>Mass = ?</span><span><span /></span>
<span><span>2. </span>Density = 1298 g/L</span>
3. Volume = 125mL to L
a. 125 ml x 0.001l/1ml = 0.125 L
<span>Formula and derivation: </span><span><span>
1. </span>density = mass / volume</span> <span><span>
2. mass </span>= density / volume</span>
<span>Solution for the problem: </span><span><span>
1. mass = </span></span> <span> 1298 g/L / 0.125 L = 10384g
</span>
Answer:
The specific heat for the metal is 0.466 J/g°C.
Explanation:
Given,
Q = 1120 Joules
mass = 12 grams
T₁ = 100°C
T₂ = 300°C
The specific heat for the metal can be calculated by using the formula
Q = (mass) (ΔT) (Cp)
ΔT = T₂ - T₁ = 300°C - 100°C = 200°C
Substituting values,
1120 = (12)(200)(Cp)
Cp = 0.466 J/g°C.
Therefore, specific heat of the metal is 0.466 J/g°C.
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.
Answer:
=759.95 grams.
Explanation:
The molar mass of chromium is 51.9961 g/mol
Therefore the number of moles of chromium in 156 grams is:
Number of moles =mass/RAM
=156g/51.9961g/mol
=3 moles.
From the equation provided, 3 moles of chromium metal produce 2 moles of Chromium oxide.
Therefore 3 moles of chromium produce:
(3×2)/4 moles =1.5 moles of chromium oxide.
I mole of chromium oxide has a mass of 151.99 g
Thus 1.5 moles= 1.5mole ×151.99 g/mol
=759.95 grams.
Answer:
S = 7.9 × 10⁻⁵ M
S' = 2.6 × 10⁻⁷ M
Explanation:
To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0
C +S +S
E S S
The solubility product (Ksp) is:
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²
S = 7.9 × 10⁻⁵ M
<u>Solubility in 0.0120 M CoBr₂ (S')</u>
First, we will consider the ionization of CoBr₂, a strong electrolyte.
CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)
1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.
Then,
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0.0240
C +S' +S'
E S' 0.0240 + S'
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')
In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.
S' = 2.6 × 10⁻⁷ M
