<h3>
Answer:</h3>
1.9 moles
<h3>
Explanation:</h3>
Carbon dioxide (CO₂) is a compound that is made up of carbon and oxygen elements.
It contains 2 moles of oxygen atoms and 1 mole of carbon atoms
Therefore;
We would say, 1 mole of CO₂ → 2 moles of Oxygen atoms + 1 mole of carbon atoms
Thus;
If a sample of CO₂ contains 3.8 moles of oxygen atoms we could use mole ratio to determine the moles of CO₂
Mole ratio of CO₂ to Oxygen is 1 : 2
Therefore;
Moles of CO₂ = 3.8 moles ÷ 2
= 1.9 moles
Hence, the moles of CO₂ present in a sample that would produce 3.8 moles of Oxygen atoms is 1.9 moles
Answer:
D. ![K_{a} = \frac{[\text{H}^{+}][\text{NO}_{2}^{-}]}{[\text{HNO}_{2}]}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%5B%5Ctext%7BNO%7D_%7B2%7D%5E%7B-%7D%5D%7D%7B%5B%5Ctext%7BHNO%7D_%7B2%7D%5D%7D)
Explanation:
The general form of an equilibrium constant expression is
![K = \frac{[\text{Products}]}{[\text{Reactants}]}](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5B%5Ctext%7BProducts%7D%5D%7D%7B%5B%5Ctext%7BReactants%7D%5D%7D)
In the equilibrium
HNO₂ ⇌ H⁺ + NO₂⁻
The products are H⁺ and NO₂⁻, and the reactant is HNO₂.
∴
<span>The reaction rate increases.
Why </span><span>Well a catalyst usually lower the activation barrier in an energy diagram. The lower and smaller that gap means the reaction is taking place rapidly compared to when that activation barrier gap is higher. </span>
Answer:
0.170 M
Explanation:
As this is a <em>series of dilutions</em>, we can continuosly<em> use the C₁V₁=C₂V₂ formula </em>to solve this problem:
For the first step:
- 59.0 mL * 1.80 M = 258 mL * C₂
Then for when 129 mL of that 0.412 M are diluted by adding 183 mL of water:
- V₂ = 129 mL + 183 mL = 312 mL
Using <em>C₁V₁=C₂V₂:</em>
- 129 mL * 0.412 M = 312 mL * C₂
