Answer:
0.15 L
Explanation:
Step 1. Calculate the <em>moles of KF.</em>
Molar mass = 58.10 g/mol
Moles of KF = 8.0 × 1/58.10
Moles of KF = 0.138 mol KF
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Step 2. Calculate the <em>volume of KF
</em>
c = n/V Multiply both sides by V
V = 0.138 × 1/0.89
V = 0.15 L
Answer:
2RbNO₃ + BeF₂ → Be(NO₃)₂ + 2RbF, because Be keeps a 2+ charge throughout the reaction
Explanation:
2RbNO₃ + BeF₂ → Be(NO₃)₂ + 2RbF, because Be keeps a 2+ charge throughout the reaction
Rb is a +1 cation, NO3 is a -1 anion, Be is a +2 cation and F is a -1 anion.
In writing an ionic compound the charge of the cation becomes the subscript of the anion and the charge of the anion becomes the subscript of the cation.
So the ionic compound formed between Be2+ and F- is BeF2. The ionic compound formed between Be2+ and NO3- is Be(NO₃)₂.
As there are two NO₃ on the product side it is balanced by writing a 2 coefficient before RbNO₃ on the reactant side.
And as there are two F on the reactant side it is balanced by writing a 2 coefficient before RbF on the product side.
Answer:
its 57.56
Explanation:
dont ask me how it is just is trust me
For every two AB produced, the reaction requires three A
Answer:
0.098 moles
Explanation:
Let y represent the number of moles present
1 mole of Ba(OH)₂ contains 2 moles of OH- ions.
Hence, 0.049 moles of Ba(OH)2 contains y moles of OH- ions.
To get the y moles, we then do cross multiplication
1 mole * y mole = 2 moles * 0.049 mole
y mole = 2 * 0.049 / 1
y mole = 0.098 moles of OH- ions.
1 mole of OH- can neutralize 1 mole of H+
Therefore, 0.098 moles of HNO₃ are present.
