In order to find the answer, use an ICE chart:
Ca(IO3)2...Ca2+......IO3-
<span>some.......0..........0 </span>
<span>less.......+x......+2x </span>
<span>less........x.........2x
</span>
<span>Ca(IO₃)₂ ⇄ Ca⁺² + 2 IO⁻³
</span>
K sp = [Ca⁺²][IO₃⁻]²
K sp = (x) (2 x)² = 4 x³
7.1 x 10⁻⁷ = 4 x³
<span>x = molar solubility = 5.6 x 10</span>⁻³ M
The answer is 5.6 x 10 ^ 3 M. (molar solubility)
The amount of sodium bromide that would be formed from the reaction will be 7.5524 grams
<h3>Stoichiometric calculation</h3>
Looking at the equation of the reaction:
The mole ratio of CuBr2 and NaCl is 1:2.
Mole of 16.4 grams of CuBr2 = 16.4/223.37
= 0.0734 moles
Mole of 22.7 grams of NaCl = 22.7/58.44
= 0.3884 moles
Equivalent mole of NaCl = 0.1468 moles
Thus, NaCl is in excess while CuBr2 is limiting.
Mole ratio of CuBr2 and NaBr = 1:1
Mass of 0.0734 mole NaBr = 0.0734 x 102.894
= 7.5524 grams
More on stoichiometric calculation can be found here: brainly.com/question/8062886
The temperature will be the average of 10 C and 20 C which is 15 C.
Energy is transferred from faster moving molecules to slower moving molecules, and the result is the average of 15°C. <span>This is just another temperature problem where you add something hot to a cold liquid and you have to calculate the final temperature. Since the energy given up by the hot coffee = energy absorbed by the cold coffee, the mass of the two fluids is the same (each is 1 cup) and both are coffee so the specific heat is the same, you get T hot - Tfinal = Tfinal - Tcold and then you solve it, you get T final = (T hot + Tcold) / 2 which is simply the average of the hot and cold temperatures.
BRAINLIEST PLS!</span>
The isotope that is more abundant, given the data is isotope Li7
<h3>Assumption</h3>
- Let Li6 be isotope A
- Let Li7 be isotope B
<h3>How to determine whiche isotope is more abundant</h3>
- Molar mass of isotope A (Li6) = 6.02 u
- Molar mass of isotope B (Li7) = 7.02 u
- Atomic mass of lithium = 6.94 u
- Abundance of A = A%
- Abundance of B = (100 - A)%
Atomic mass = [(mass of A × A%) / 100] + [(mass of B × B%) / 100]
6.94 = [(6.02 × A%) / 100] + [(7.02 × (100 - A)) / 100]
6.94 = [6.02A% / 100] + [702 - 7.02A% / 100]
6.94 = [6.02A% + 702 - 7.02A%] / 100
Cross multiply
6.02A% + 702 - 7.02A% = 6.94 × 100
6.02A% + 702 - 7.02A% = 694
Collect like terms
6.02A% - 7.02A% = 694 - 702
-A% = -8
A% = 8%
Thus,
Abundance of B = (100 - A)%
Abundance of B = (100 - 8)%
Abundance of B = 92%
SUMMARY
- Abundance of A (Li6) = 8%
- Abundance of B (Li7) = 92%
From the above, isotope Li7 is more abundant.
Learn more about isotope:
brainly.com/question/24311846
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Answer:
A salt bridge or ion bridge, in electrochemistry, is a laboratory device used to connect the oxidation and reduction half-cells of a galvanic cell (voltaic cell), a type of electrochemical cell. It maintains electrical neutrality within the internal circuit.
Explanation:
