Answer:
bonding driven by ionic interactions.
Explanation:
The answer for this question is 0.8
<h3>
Answer:</h3>
2.125 g
<h3>
Explanation:</h3>
We have;
- Mass of NaBr sample is 11.97 g
- % composition by mass of Na in the sample is 22.34%
We are required to determine the mass of 9.51 g of a NaBr sample.
- Based on the law of of constant composition, a given sample of a compound will always contain the sample percentage composition of a given element.
In this case,
- A sample of 11.97 g of NaBr contains 22.34% of Na by mass
A sample of 9.51 g of NaBr will also contain 22.345 of Na by mass
% composition of an element by mass = (Mass of element ÷ mass of the compound) × 100
Mass of the element = (% composition of an element × mass of the compound) ÷ 100
Therefore;
Mass of sodium = (22.34% × 9.51 g) ÷ 100
= 2.125 g
Thus, the mass of sodium in 9.51 g of NaBr is 2.125 g
Autoionization Reactions are those reactions in which ions or molecules ionizes spontaneously without adding any external reagent.
For Example,
Autoionization of water.
H₂O + H₂O ⇆ H₃O⁺ + OH⁻
Autoionization reaction of Methanol is shown below,
Answer:

Explanation:
Hello!
In this case, since we are considering an gas, which can be considered as idea, we can write the ideal gas equation in order to write it in terms of density rather than moles and volume:

Whereas MM is the molar mass of the gas. Now, since we can identify the initial and final states, we can cancel out R and MM since they remain the same:

It means we can compute the final density as shown below:

Now, we plug in to obtain:

Regards!