Answer:
Charge= -2.
Gains two electron into the 4p^4 to become 4p^6.
Explanation:
The element in the periodic table/chart that matches with the valence electron configuration is Selenium with full electron configuration of [Ar] 3d^10 4s^2 4P^4 which is a non-metal that is found in group 4 of the periodic table/chart.
Selenium can receive 2 more electrons on the 4p^4 to give a -2(minus 2) ion that is Se^2-.
Selenium can also loose 2 electron from 4s^2 to give a +2 ion that is Se^2+.
Selenium can also loose 2 electrons from 4s^2 and 2 electrons from 4p^4 to form Se^4+.
Selenium can also loose 2 electrons from 4s^2 and 4 electrons from 4p^4 to form Se^6+.
Thus, in order to form a monatomic ion with a charge(we will be making use of the most stable one). Thus, it will gain two more electron since this is easier to become 4s^2 4p^6.
Answer:
2
Explanation:
First, find the hydronium ion concentration of the solution with a pH of 4.
[H₃O⁺] = 10^-pH
[H₃O⁺] = 10⁻⁴
[H₃O⁺] = 1 × 10⁻⁴
Next, multiple the hydronium ion concentration by 100 to find the hydronium ion concentration of the new solution.
[H₃O⁺] = 1.0 × 10⁻⁴ × 100 = 0.01
Lastly, find the pH.
pH = -log [H₃O⁺]
pH = -log (0.01)
pH = 2
The pH of a solution that has a hydronium ion concentration 100 times greater than a solution with a pH of 4 is 2.
Hope this helps.
Answer:
- <em>The average mass of calcium in each sample is: </em><u>0.978 g</u>
<em />
- <em>The absolute uncertainty is: </em><u>0.008 g</u>
Explanation:
The <em>absolute uncertainty </em>of the total samples indicated in the statement is ± 0.1 g.
When you multiply or divide quantities with uncertainties, you calculate the final uncertanty by adding the <em>relative uncertainties</em> together.
The relative uncertainty is the absolute uncertainty divided by the quantity:
- Relative uncertainty = 0.1g / 12.2 g = 0.008
The average mass of calcium is calculated using proportions, along with the molar masses:
- Molar mass of calcium: 40.078 g/ mol (from a periodic table)
- Molar mass of calcite: 100.085 g/mol (given)
Proportion:
- 40.078 g of calcium / 100.085 g of calcite = x / 12.2 g of calcite
- x = 12.2 × 40.078 / 100.085 g = 4.89 g calcium
So the total mass of calcium in the five samples is 4.89 g, and the average mass in each sample is:
- Average mass = total mass of five samples / number of samples
- Average mass = 4.89 g / 5 = <u>0.978 g of calcium</u>
So, the first answer is that the average mass of calcium in each sample is 0.978 g ( keep 3 signficant figures, such as the quntitiy 12.2 shows, as you have only used multiplication and division).
The absolute uncertainty of each sample is the relative uncertainty multiplied by the average mass of calcium of the five samples, rounded to one decimal:
- Absolute uncertainty = 0.978 g × 0.008 ≈ 0.008 g
The answer to the secon question is that the absolute uncertaingy of calcium in each sample is 0.008 g.
