Answer:
The solubility of MnS will decrease on addition of KOH solution.
Explanation:
As per the equation given:

On dissolution of MnS in water it gives a basic solution as it gives hydroxide ions.
Now when the we are adding aqueous KOH solution, it will dissociate as:

Thus it will further furnish more hydroxide ion,
This will increase the concentration of hydroxide ions (present of product side), the system will try to decrease its concentration by shifting towards reactant side.
Thus the solubility of MnS will decrease on addition of KOH solution.
Answer:
The average atomic mass of bromine is 79.9 amu.
Explanation:
Given data:
Percentage of Br⁷⁹ = 55%
Percentage of Br⁸¹ = 45%
Average atomic mass of bromine = ?
Formula:
Average atomic mass = [mass of isotope× its abundance] + [mass of isotope× its abundance] +...[ ] / 100
Now we will put the values in formula.
Average atomic mass = [55 × 79] + [81 ×45] / 100
Average atomic mass = 4345 + 3645 / 100
Average atomic mass = 7990 / 100
Average atomic mass = 79.9 amu
The average atomic mass of bromine is 79.9 amu.
Problem One
You will use both m * c * deltaT and H = m * heat of fusion.
Givens
m = 12.4 grams
c = 0.1291
t1 = 26oC
t2 = 1204
heat of fusion (H_f) = 63.5 J/grams.
Equation
H = m * c * deltaT + m * H_f
Solution
H = 12.4 * 0.1291 * (1063 - 26) + 12.4 * 63.5
H = 1660.1 + 787.4
H = 2447.5 or 2447.47 is the exact answer. I have to leave the rounding to you. I have no idea where to round it although I suspect 2450 would be right for 3 sig digs.
Problem Two
Formula and Givens
t1 = 14.5
t2 = 50.0
E = 5680
c = 4.186
m = ??
E = m c * deltaT
Solution
5680 = m * 4.186 * (50 - 14.5)
5680 = m * 4.186 * (35.5)
5680 = m * 148.603 * m
m = 5680 / 148.603
m = 38.22 grams That isn't very much. Be very sure you are working in joules. You'd leave that many grams in the kettle after drying it thoroughly.
m = 38.2 to 3 sig digs.
Answer:
22.5mL is the volume of the water
Explanation:
When the graduated cylinder is in the 25.0mL mark, the mass of this volume is 22.4g. To convert this mass to volume we need to use density, as follows:
22.4g × (1mL / 0.99704g) = 22.5mL is the volume of the water.
That means the cylinder is uncalibrated in 2.5mL when the cylinder is in the 25.0mL mark
I believe the answer is MgCl, this is because the bases of group one elements (alkali metals) such as sodium, lithium , and cesium, are all strong bases unlike those of group 2 and group 3. The bases of group one elements ( NaOH, LiOH, CsOH) all readily dissolve in water forming strong soluble base. Then between a base of Ba (Ba(OH)2)and that of Mg, (Mg(OH)2), the one for Ba is more stronger because Ba is more reactive than that of Mg because it is not readily dissolve in water ( less soluble in water to give OH- ions)