Argon has 24 known isotopes.
The mass of carbon contained in 2.25 g of potassium carbonate, K₂CO₃ is 0.196 g.
<h3>
Molecular mass of potassium carbonate</h3>
The molecular mass of potassium carbonate, K₂CO₃ is calculated as follows;
M = K₂CO₃
M = (39 x 2) + (12) + (16 x 3)
M = 138 g
mass of carbon in potassium carbonate, K₂CO₃ is = 12 g
The mass of carbon contained in 2.25 g of potassium carbonate, K₂CO₃ is calculated as follows;
138 g ------------ 12 g of carbon
2.25 g ------------ ?
= (2.25 x 12) / 138
= 0.196 g
Thus, the mass of carbon contained in 2.25 g of potassium carbonate, K₂CO₃ is 0.196 g.
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In NaMnO₄, Mn has the highest oxidation number.
The question is incomplete, the complete question is;
Which of the following species contains manganese with the highest oxidation number?
A) Mn
B) MnF₂
C) Mn₃(PO₄)₂
D) MnCl₄
E) NaMnO₄
In order to ascertain the specie that contains manganese with the highest oxidation number, we must calculate the oxidation number of manganese in each of the species one after the other.
1) For Mn, the oxidation number of Mn is zero because the atom is uncombined.
2) For MnF₂;
Mn has an oxidation number of +2
3) For Mn₃(PO₄)₂
Mn has an oxidation number of +2
4) For MnCl₄
Mn has an oxidation number of +4
5) For NaMnO₄
Mn has an oxidation number of +7
Hence in NaMnO₄, Mn has the highest oxidation number.
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Answer:
2Na⁺ (aq) and 2OH⁻(aq)
Explanation:
Spectator ions:
Spectator ions are those ions which are same on both side of chemical reaction. These ions are same in the reactant side and product side. Their presence can not effect the chemical equilibrium that's why when we write the net ionic equation these ions are neglect or omitted.
Given ionic equation:
Ba⁺²(aq) + 2OH⁻(aq) + 2Na⁺ (aq) + CO²⁻₃(aq) → BaCO₃(s) + + 2Na⁺ (aq) + 2OH⁻(aq)
In given ionic equation by omitting the spectator ions i.e, 2Na⁺ (aq) and 2OH⁻(aq) net ionic equation can be written as,
Net ionic equation:
Ba⁺²(aq) + CO²⁻₃(aq) → BaCO₃(s)
Answer:
0.0432M
Explanation:
We begin by writing a balanced equation for the reaction. This is illustrated below:
NaOH + HCl —> NaCl + H2O
From the equation above,
The number of mole of the acid (nA) = 1
The number of mole of the base (nB) = 1
Data obtained from the question include:
Vb (volume of the base) = 54mL
Cb (concentration of the base) = 0.1M
Va (volume of the acid) = 125mL
Ca ( concentration of the acid) =?
Using CaVa/CbVb = nA/nB, the concentration of the acid can easily be obtained as shown below:
CaVa/CbVb = nA/nB
Ca x 125 / 0.1 x 54 = 1
Cross multiply to express in linear form:
Ca x 125 = 0.1 x 54
Divide both side by 125
Ca = (0.1 x 54) / 125
Ca = 0.0432M
Therefore, the concentration of the acid is 0.0432M