Explanation:
Let us assume that the given data is as follows.
mass of barium acetate = 2.19 g
volume = 150 ml = 0.150 L (as 1 L = 1000 ml)
concentration of the aqueous solution = 0.10 M
Therefore, the reaction equation will be as follows.
Hence, moles of = .......... (1)
As, No. of moles =
Hence, moles of will be calculated as follows.
No. of moles =
= (molar mass of is 255.415 g/mol)
=
Moles of =
= 0.01715 mol
Hence, final molarity will be as follows.
Molarity =
=
= 0.114 M
Thus, we can conclude that final molarity of barium cation in the solution is 0.114 M.
Answer:
It will turn red
Explanation:
Bases will turn the litmus paper (impregnated with an acid-base indicator) blue.
Acids will turn turn the litmus paper (impregnated with an acid-base indicator) red.
Since the produced HCl is a strong acid, the litmus paper will turn red, when touching the HCl.
The red shows the presence of an acid, in this case HCl.
Answer: No reaction occurs
Explanation:.
Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.
The given chemical equation is:
The complete ionic equation is;
The ions which are present on both the sides of the equation are are not involved in net ionic equation.
Hence, there is no net reaction.
Answer is: D. Na2SO4.
b(solution) = 0.500 mol ÷ 2.0 L.
b(solution) = 0.250 mol/L.
b(solution) = 0.250 m; molality of the solutions.
ΔT = Kf · b(solution) · i.
Kf - the freezing point depression constant.
i - Van 't Hoff factor.
Dissociation of sodium sulfate in water: Na₂SO₄(aq) → 2Na⁺(aq) + SO₄²⁻(aq).
Sodium sulfate dissociates on sodium cations and sulfate anion, sodium sulfate has approximately i = 3.
Sodium chloride (NaCl) and potassium iodide (KI) have Van 't Hoff factor approximately i = 2.
Carbon dioxide (CO₂) has covalent bonds (i = 1, do not dissociate on ions).
Because molality and the freezing point depression constant are constant, greatest freezing point lowering is solution with highest Van 't Hoff factor.
Answer:
0.1113 mol
Explanation:
Data Given:
no. of atoms of CH₄= 6.70 x 10²² atoms
no. of moles of methane (CH₄) = ?
Solution:
we will find no. of moles of methane (CH₄)
Formula used
no. of moles = no. of atoms / Avogadro's number
Where
Avogadro's number = 6.022 x 10²³
Put values in above equation
no. of moles = 6.70 x 10²² atoms / 6.022 x 10²³ (atoms/mol)
no. of moles = 0.1113 mol
So,
There are 0.1113 moles of methane.