How many atoms of K are present in 195.49 grams of K? 3.0110 × 1024 6.0220 × 1024 1.1772 × 1026 4.5797 × 1027
1 answer:
<h2>Hello!</h2>
The answer is:
First, we need to look for the K molecular mass:
Also, we need to remember that 1 mol of a compound or element is equal to of that compound.
From the statement, we are asked to calculate how many atoms of K are present in 195,49 of K.
So, doing the calculations, we have:
So, we have atoms of K in 195,49 grams of K.
Have a nice day!
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Answer:
pH = 9,57
[M²⁻] = 7,948x10⁻²M
[HM⁻] = 4x10⁻⁵M
[H₂M] = 0M
Explanation:
The moles of maleic acid presents in the solution are:
0,923g×=7,952x10⁻³moles of H₂M
60,0mL of 0,265M KOH are:
0,0600L×=0,0159 moles of KOH
The reactions of maleic acid (H₂M) and then with HM⁻ are:
H₂M + KOH → HM⁻ + H₂O + K⁺ (1)
HM⁻ + KOH → M²⁻ + H₂O + K⁺ (2)
For a complete transformation of H₂M in HM⁻ there are necessaries 7,952x10⁻³moles of KOH. As the moles of KOH are 0,0159 moles, the restant moles are:
0,0159 - 7,952x10⁻³ = <em>7,948x10⁻³ moles of KOH</em>
By (2), the moles produced of M²⁻ are the same as moles of KOH, <em>7,948x10⁻³ moles, </em>and moles of HM⁻ are:
7,952x10⁻³ -<em> </em>7,948x10⁻³ <em> = 4x10⁻⁶ moles of HM⁻</em>
Using Henderson-Hasselbalch formula:
pH = pka + log₁₀ [M²⁻] /[HM⁻]
pH = 6,27 + log₁₀ <em>7,948x10⁻³ / 4x10⁻⁶</em>
pH = 9,57
The moles of M²⁻ are 7,948x10⁻³ and volume of the solution is 0,1000L,
[M²⁻] = 7,948x10⁻²M
Moles of HM⁻ are 4x10⁻⁶:
[HM⁻] = 4x10⁻⁵M
And there is not H₂M:
[H₂M] = 0M
I hope it helps!