<u>Answer:</u> The amount of sodium thiosulfate required is moles
<u>Explanation:</u>
Moles of solution given = 0.0000524 moles
The chemical equation for the reaction of potassium iodate and sodium thiosulfate follows:
By Stoichiometry of the reaction:
2 moles of potassium iodate reacts with 1 mole of sodium thiosulfate
So, 0.0000524 moles of potassium iodate will react with = of sodium thiosulfate
Hence, the amount of sodium thiosulfate required is moles
It is typically formed by the evaporation of salty water (such as sea water) which contains dissolved Na+ and Cl- ions. One finds rock salt deposits ringing dry lake beds, inland marginal seas, and enclosed bays and estuaries in arid regions of the world
Answer:
14,250 years old.
Explanation:
First, multiply the two lifetimes. - 11,400.
Then, divide 5,700 in half for the 0.5 lifetime. - 2,850.
Lastly, add them all up. - 11,400 + 2,850 = 14,250 years old.
sorry if incorrect <3
thats what i understood XD
Answer:
BaI₂.5H₂O
Explanation:
Given Data:
Mass of Hydrated BaI₂ = 10.222 g
Mass of dried BaI₂ = 9.520 g
Mass of Water removed = 10.222-9.520 = 0.702 g
M.Mass of BaI₂ = 391.136 g/mol
M.Mass of Water = 18.02 g/mol
Now,
Calculate moles of dried BaI₂ as,
Moles = Mass / M.Mass
Moles = 9.520 g / 391.136 g/mol
Moles = 0.02434 moles
Calculate moles of Water as,
Moles = Mass / M.Mass
Moles = 0.702 g / 18.02 g/mol
Moles = 0.0389 moles
Then,
Calculate Mole ratio of BaI₂ and water as,
= 0.02434 moles BaI₂ / 0.0389 moles Water
= 0.625
Now,
We will convert this mole ratio to a whole number by multiplying it with a nearest integer,
= 0.625 × 8
= 5
Hence, this means for every one mole of BaI there are 5 moles of Water.
Result:
BaI₂.5H₂O
Please, you have to apply the formula below:<span>Q=c∗m∗Δt</span>where Q is the energy lost, c is the specific heat of water, m is the mass of water involved, so m=3.75 *10^-1 Kg c=4,184 J/(Kg*°C) delta t=37.5 °C
Taking density of water as 1000kg/m3. Mass of water would be 0.375kg. So, heat lost would be<span>H=mCDeltaT</span>H=0.375*4184*37.5 = 58837.5J