Answer:
2 mol H
Explanation:
For every 2 mol of NaOH, we're reacting 2 mol of H2O. In order to figure out how many mol of H are needed, it needs to be set up stochiometrically. Starting off with the given value, 1 mol of NaOH, we can then make a mol to mol ratio. For 2 mol of NaOH, we have 2 mol of H2O. For every 2 mol of H2O, we have 4 mol of H (this is because we are multiplying the coefficient by the subscript: 2 × 2). Now, we can solve for our answer.
1 mol NaOH × (2 mol H₂O / 2 mol NaOH) × (4 mol H / 2 mol H₂O)
= 2 mol H
Thus, we get 2 mol of H are needed to completely react 1 mol of NaOH.
In a fully inflated basketball, there are more particles of gas.
Answer: Option D.
<u>Explanation:</u>
When a ball is fully inflated, it means it has a lot of air in it. With a lot of air in it, it means that there a lot of gas particles in that ball. With the large amount of gas particles in it, there is a lot of collision among these particles inside the ball.
This collision leads to the pressure on the greater side compared to the ball which has less air in it and is deflated because it has less particles of gas inside.
It just is because science is science and it's supposed to make sense
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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The way you want to find the percent composition would be by breaking down the problem like so:
K= atomic mass of K which is 39.098
Mn = atomic mass of Mn which is 54.938
O= atomic mass of o which is 15.999
Then you want to add 39.098+ 54.938+ 15.999 and you get 110.035 which is the molar mass for KMnO
Then you want to take each molar mass and then divide it 110.035 and multiply by 100
Ex. K = 39.098/ 110.035 and the multiply what you get by a 100
You do this for the other elements as well good luck!
