<h3>
Answer:</h3>
70.906 g
<h3>
Explanation:</h3>
We are given;
- Atoms of Chlorine = 1.2 × 10^24 atoms
We are required to calculate the mass of Chlorine
- We know that 1 mole of an element contains atoms equivalent to the Avogadro's number, 6.022 × 10^23.
- That is , 1 mole of an element = 6.022 × 10^23 atoms
- Therefore; 1 mole of Chlorine = 6.022 × 10^23 atoms
But since Chlorine gas is a molecule;
- 1 mole of Chlorine gas = 2 × 6.022 × 10^23 atoms
But, molar mass of Chlorine gas = 70.906 g/mol
Then;
70.906 g Of chlorine gas = 2 × 6.022 × 10^23 atoms
= 1.20 × 10^24 atoms
Thus;
For 1.2 × 10^24 atoms ;
= ( 70.906 g/mol × 1.2 × 10^24 atoms ) ÷ (1.20 × 10^24 atoms)
<h3>= 70.906 g </h3>
Therefore, 1.20 × 10^24 atoms of chlorine contains a mass of 70.906 g
=
Answer:
This is a coal combustion process and we will assume
Inlet coal amount = 100kg
It means that there are
15kg of H2O, 2kg of Sulphur and 83kg of Carbon
Now to find the mole fraction of SO2(g) in the exhaust?
Molar mass of S = 32kg/kmol
Initial moles n of S = 2/32 = 0.0625kmols
Reaction: S + O₂ = SO₂
That is 1 mole of S reacts with 1 mole of O₂ to give 1 mole of SO₂
Then, it means for 0.0625 kmoles of S, we will have 0.0625 kmole of SO2 coming out of the exhaust
The mole fraction of SO2(g) in the exhaust=0.0625kmols
Explanation:
Answer: Your answer would be D. A neutral atom would become positively charged when they lose electrons because electrons are negatively charged. Removing some of these negative charges causes the atom to become more positively charged. Hope this has helped you :D
Answer:
Answer choices A, B, and C all contain a certain level <em>(or amount, if you will)</em> of <u>Calcium Carbonate</u>, but <u><em>B. chalk</em></u> contains the most amount out of all three.
Explanation:
I hope that this was helpful! (✿◠‿◠)
Thank you!
Answer:
Universal indicator can show us how strongly acidic or alkaline a solution is, not just that the solution is acidic or alkaline. This is measured using the pH scale , which runs from pH 0 to pH 14.
Explanation:
~Hope this helps