Answer:
Lake and River pollution?
Explanation:
Bohr model is valid only for hydrogen and hydrogen-like species, but quantum mechanical model can explain all elements....
Answer:
D.
Explanation:
INCREASE OF SOLAR WINDS When the sun is more active
Answer:

Explanation:
We are given a number of particles and asked to convert to moles.
<h3>1. Convert Particles to Moles </h3>
1 mole of any substance contains the same number of particles (atoms, molecules, formula units) : 6.022 *10²³ or Avogadro's Number. For this question, the particles are not specified.
So, we know that 1 mole of this substance contains 6.022 *10²³ particles. Let's set up a ratio.

We are converting 2.98*10²³ particles to moles, so we multiply the ratio by that value.

The units of particles cancel.



<h3>2. Round</h3>
The original measurement of particles (2.98*10²³) has 3 significant figures, so our answer must have the same.
For the number we found, 3 sig figs is the thousandth place.
The 8 in the ten-thousandth place (0.4948522086) tells us to round the 4 up to a 5 in the thousandth place.

2.98*10²³ particles are equal to approximately <u>0.495 moles.</u>
Answer:
0.2788 M
1.674 %(m/V)
Explanation:
Step 1: Write the balanced equation
NaOH + CH₃COOH → CH₃COONa + H₂O
Step 2: Calculate the reacting moles of NaOH

Step 3: Calculate the reacting moles of CH₃COOH
The molar ratio of NaOH to CH₃COOH is 1:1.

Step 4: Calculate the molarity of the acetic acid solution

Step 5: Calculate the mass of acetic acid
The molar mass of acetic acid is 60.05 g/mol.

Step 6: Calculate the percentage of acetic acid in the solution
