Answer:
0.32M
Explanation:
<u>Step 1:</u> Balance the reaction
K2CO3 + Ba(NO3)2 ⇔ KNO3 + BaCO3
We have a 20 mL 0.2 M K2CO3 and a 30mL 0.4M Ba(NO3)2 solution
SinceK2CO3 is the limiting reactant, there will remain Ba(NO3)2 after it's consumed and produced KNO3 + BaCO3
<u>Step 2: </u>Calculate concentration
To find the concentration of the barium cation we use the following equation:
Concentration = moles of the <u>solute</u> / volumen of the <u>solution</u>
<u />
<u>[Ba2+] </u> = (20 * 10^-3 * 0.2M + 30 * 10^-3 * 0.4M) / ( 20 + 30mL) *10^-3
[Ba2+] = 0.32 M
The concentration of Barium ion in solution is 0.32 M
I'm going to go with Physical.
C. It is decreased by a factor of 3.
It seems that you have missed the given image to answer this question. But anyway, I found it and got the answer. Based on the topographical map of a section of Charleston, SC, the feature that is <span>located at the dot marked with an X is the high point of a hill. The answer would be option D.</span>
The total number of ions in 38.1 g of SrF₂ is 5.479 x 10²³.
<h3>What are ions?</h3>
Ions are the elements with a charge on them. It happens when they share electrons with other atoms to form a compound.
We have to calculate the total number of ions in 38.1 g of .
The molar mass of SrF₂ = 125.62 g/mol
The number of moles = 38.1 g of 1.0 mol / 125.62 = 0.30329 moles
Given that, total moles of SrF₂ ions in = 1.0 mol of + 2.0 moles of = 3.0 moles
Total moles of ions in 0.30329 moles of
= (0.30329 moles of SrF₂) x 3.0 / 1.0 = 0.90988 mol ions
We know that,
1.0 mole of ions = 6.023 x 10²³ ions
Thus, the number of total ions = ( 0.90988 mol ions) x 6.023 x 10²³ / 1.0 mol = 5.479 x 10²³ ions
Thus, the number of ions is in 38.1 g of 5.479 x 10²³ ions
To learn more about ions, refer to the link:
brainly.com/question/14295820
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