Answer:
Concentration solution A was 0.5225 M
Explanation:
10.00 mL of solution A was diluted to 50.00 mL and yields 50.00 mL of solution B
According to laws of dilution- 
where, 
and 
are concentration of solution A and B respectively
and 
are volumes of solution A and B respectively
Here = 0.1045 M, = 50.00 mL and = 10.00 mL
Hence, 
So, concentration solution A was 0.5225 M
Answer:
The formula of the original halide is SrCl₂.
Explanation:
- The balanced equation of this reaction is:
SrX₂ + H₂SO₄ → SrSO₄ + 2 HX, where X is the halide.
- From the equation stichiometry, 1.0 mole of strontium halide will result in 1.0 mole of SrSO₄.
- The number of moles of SrSO₄ <em>(n = mass/molar mass) </em>= (0.755 g) / (183.68 g/mole) = 4.11 x 10⁻³ mole.
- The number of moles of SrX are 4.11 x 10⁻³ moles from the stichiometry of the balanced equation.
- n = mass / molar mass, n = 4.11 x 10⁻³ moles and mass = 0.652 g.
- The molar mass of SrX₂ = mass / n = (0.652) / (4.11 x 10⁻³ moles) = 158.62 g/mole.
- The molar mass of SrX₂ (158.62 g/mole) = Atomic mass of Sr (87.62 g/mole) + (2 x Atomic mass of halide X).
- The atomic mass of halide X = (158.62 g/mole) - (87.62 g/mole) / 2 = 71 / 2 g/mole = 35.5 g/mole.
- This is the atomic mass of Cl.
- <em>So, the formula of the original halide is SrCl₂</em>.
Answer:
Temperature is an abiotic component of an ecosystem
Explanation:
Answer:
2Ag(s) + Cu^2+(aq) ----------> 2Ag^+(aq) + Cu(s)
Explanation:
Ag(s)/Ag^+ (aq) is the anode as shown while Cu^2+(aq)/Cu^2(s) is the cathode.
E°cell= E°cathode -E°anode= 0.34 -0.80= -0.5V
The cell is not spontaneous as written because E°cell is negative. This implies that the electrodes of the cell must be interchanged to make the cell spontaneous.
The equilibrium expression shows the ratio between
products and reactants. This expression is equal to the concentration of the
products raised to its coefficient divided by the concentration of the
reactants raised to its coefficient. The correct equilibrium expression for the
given reaction is:<span>
<span>H2CO3(aq) + H2O(l) = H3O+(aq) + HCO3-1(aq)
Kc = [HCO3-1] [H3O+] / [H2O] [H2CO3]</span></span>
