Answer:
The mass of
4.6
×
10
24
atoms of silver is approximately 820 g.
Explanation:
In order to determine the mass of a given number of atoms of an element, identify the equalities between moles of the element and atoms of the element, and between moles of the element and its molar mass.
1
mole atoms Ag=6.022xx10
23
atoms Ag
Molar mass of Ag =#"107.87 g/mol"#
Multiply the given atoms of silver by
1
mol Ag
6.022
×
23
atoms Ag
. Then multiply times the molar mass of silver.
4.6
×
10
24
atoms Ag
×
1
mol Ag
6.022
×
10
23
atoms Ag
×
107.87
g Ag
1
mol Ag
=
820 g Ag
The way you want to find the percent composition would be by breaking down the problem like so:
K= atomic mass of K which is 39.098
Mn = atomic mass of Mn which is 54.938
O= atomic mass of o which is 15.999
Then you want to add 39.098+ 54.938+ 15.999 and you get 110.035 which is the molar mass for KMnO
Then you want to take each molar mass and then divide it 110.035 and multiply by 100
Ex. K = 39.098/ 110.035 and the multiply what you get by a 100
You do this for the other elements as well good luck!
Answer:
T2= 7.3°C
Explanation:
To solve this problem we will use Charles law equation i.e,
V1/T1 = V2/T2
Given data
V1 = 269.7 L
T1 = 6.12 °C
V2= 320.4 L
T2=?
Solution:
Now we will put the values in equation
269.7 L / 6.12°C = 320.4 L / T2
T2= 320.4 L × 6.12°C/ 269.7 L
T2= 1960.85 °C. L /269.7 L
T2= 7.3°C
The HCl added = 1.25 moles
and the moles of Na2HPO4 = 1 mole
Now when acid is added in the given solution of Na2HPO4
One mole of H+ will react with one mole of Na2HPO4 to given one mole of NaH2PO4
Na2HPO4 + H+ ---> NaH2PO4
Now this one mole formed NaH2PO4 will further react with 0.25 moles of H+ left to form 0.25 moles of H3PO4 and 0.75 moles of NaH2PO4 will remain in the solution
So this will result into formation of a buffer of phosphoric acid and NaH2PO4
NaH2PO4 + H+ ---> H3PO4
pKa of H3PO4 = 2.1
so pH = pKa + log [salt] / [acid] = 2.1 + log [0.75 / 0.25] = 2.58
so the pH will be in between 2.1 to 7.2
Answer:
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