When we balance the given equation
SF₆(g) + SO₃(g) → SO₂F₂(g)
We will get
SF₆(g) + 2SO₃(g) → 3SO₂F₂(g)
Solution:
Balancing the given equaation
SF₆(g) + SO₃(g) → SO₂F₂(g)
We have to balance the given number of O
SF₆(g) + 2SO₃(g) → 3SO₂F₂(g)
We get balanced equation
SF₆(g) + 2SO₃(g) → 3SO₂F₂(g)
The reaction quotient will be
Qc = [product] / [reactant]
Qc = [SO₂F₂(g)] / [SF₆(g) + SO₃(g)]
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<span>You need to have NAD+ as a source of oxidation for the pyruvate, as well as a supply of coenzyme A. CO2 is released by the pyruvate as a carboxyl group is removed</span>
Explanation:
As density is defined as the mass of a substance divided by its volume.
Mathematically, Density = 
It is given that mass is 50 g and density is 0.934
.
Hence, calculate the volume of methyl acetate as follows.
Density = 
0.934
= 
Volume = 
or, =
(as 1
= 1 mL)
Thus, we can conclude that the volume of methyl acetate the student should pour out is
.
The answer to this question would be: 2 mol
To answer this question, you need to know the molecular weight of Potassium. Molecular weight determines how much the weight of 1 mol of a molecule has.
Potassium or Kalium molecular weight is 39.1 gram/mol. Then, 78.20gram of potassium should be: 78.20g/ (39.1g/mol)= 2 mol
Interphase is not a phase of mitosis.