When beryllium ion combines with carbonate ion, the numbers of each ion are??
1 answer:
Answer:
One of each
Explanation:
Be is in Group 2, so it loses its two valence electrons in a reaction to form Be²⁺ ions.
Carbonate ion has the formula CO₃²⁻.
We can use the criss-cross method to work out the formula of beryllium carbonate.
The steps are
Write the symbols of the anion and cation.
Criss-cross the numbers of the charges to become the subscripts of the other ion.
Write the formula with the new subscripts.
Divide the subscripts by their highest common factor.
Omit all subscripts that are 1.
When you use this method with Be²⁺ and CO₃²⁻, you might be tempted to write the formula for the beryllium carbonate as Be₂(CO₃)₂
However, you can divide the subscripts by their largest common factor (2).
This gives you the formula Be₁(CO₃)₁.
We omit subscripts that are 1, so the correct formula is
BeCO₃
There is one Be²⁺ ion and one CO₃²⁻ ion in a formula unit of beryllium carbonate.
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Answer:
a) The theoretical yield is 408.45g of
b) Percent yield =
Explanation:
1. First determine the numer of moles of and
.
Molarity is expressed as:
M=
- For the
M=
Therefore there are 1.75 moles of
- For the
M=}{1Lsolution}[/tex]
Therefore there are 2.0 moles of
2. Write the balanced chemical equation for the synthesis of the barium white pigment, :
3. Determine the limiting reagent.
To determine the limiting reagent divide the number of moles by the stoichiometric coefficient of each compound:
- For the :
- For the :
As the is the smalles quantity, this is the limiting reagent.
4. Calculate the mass in grams of the barium white pigment produced from the limiting reagent.
5. The percent yield for your synthesis of the barium white pigment will be calculated using the following equation:
Percent yield =
Percent yield =
The real yield is the quantity of barium white pigment you obtained in the laboratory.
Answer: The concentration of is 0.234 M
Explanation:
According to the neutralization law,
where,
= basicity
= 2
= molarity of
solution = ?
= volume of
solution = 50.0 ml
= acidity of
= 1
= molarity of
solution = 0.375 M
= volume of
solution = 62.5 ml
Putting in the values we get:
Therefore concentration of is 0.234 M