Answer:
to calculate the molarity of the said sucrose,
firstly calculate the moles
which is = Molecular weight of C12H22O11 = 342g/mol
then
moles = 139/342
= 0.41 moles
to calculate Molarity now
Molarity= moles of the solute/volume of solution in liter
=0.41/2.60
=0.158M
Explanation:
<h2>Answer:</h2>
<em>8.67kJ/mol</em>
<h2>Explanations</h2>
The formula for calculating the amount of heat absorbed by the water is given as:
Determine the moles of KI
Since heat is lost, hence the enthalpy change of the solution will be negative that is:
Determine the enthalpy of solution in kJ•mol-1
Hence the enthalpy of solution in kJ•mol-1 for KI is 8.67kJ/mol
Formula:
0.57 liters = 1 breath(17 breaths = 1minute)(60 minutes = 1 hour)(8 hours) = 4651.2 liters
each breath is 0.57 liters
He or she exhales 17(0.57) = 9.69 liters in a minute
in an hour, he or she exhales 9.69(60) liters in a minute = 581.4 liters
Answer = in 8 hours he or she exhales, 581.4(8) = 4651.2 liters in 8 hours
Hope that answers your question!!!!have a great day!!!
We have to first write a balanced equation.
so2 + o2 -> so3
this is not balanced though. we have 3 oxygen on right and 4 on left
2so2 + o2 -> 2so3
now it is same on both sides. we have to figure out which is limiting reagent with the given amounts of reagents. we do this by comparing the ratio between them in terms of moles. we see that so2 has a coefficient of 2 and o2 has none which implies 1 and so3 has 2. this means that for every 2 moles of so2 reacting with 1 mole of o2, we get 2 moles of so3.
lets convert the given values to moles. to do this we know that molecular weight is measured in grams per mole. we are given grams and need to cancel out the grams to get moles. so the molecular weight:
so2 =32.1 + 2 * 16 = 64.1 g/mol
o2 = 2 * 16 = 32 g/mol
so3 = 32.1 + 3 * 16 = 80.1 g/mol
now to convert 90 g of 2so2 under ideal conditions.
90g / 64.1g/mol = 1.404 moles
convert this amount of moles of so2 to moles of o2. we have 2 moles of so2 to 1 of o2
1.404moles so2 / 2 moles so2 * 1 mole o2= 0.702 moles o2
so we see under ideal conditions that 90g of so2 would react with .702g of o2. lets see how many we actually have with 100g of o2
100g / 32g/mol =3.16 mol.
so we have a lot more o2 than needed. we are looking for how much is left in grams. we have to figure out how much was used. to do this convert our ideal moles of o2 into grams.
.702 moles o2 * 32g/mol = 22.5g o2
so what we startrd with (100g) minus what we needed (22.5g) is what we have left
100 - 22.5 = 77.5g o2
Answer:
104.352°C
Explanation:
Data Given:
Boiling point of water = 100.0°C
Kb (boiling point constant = 0.512°C/m
Concentration of the Mg₃(PO₄)₂ = 8.5 m
Solution:
Formula Used to find out boiling point
ΔTb = m.Kb . . . . . . (1)
where
ΔTb = boiling point of solution - boiling point of water
So,
we can write equation 1 as under
ΔTb = Tb (Solution) -Tb (water)
As we have to find out boiling point so rearrange the above equation
Tb (Solution) = m.Kb + Tb (water) . . . . . . . (2)
Put values in Equation 2
Tb (Solution) = (8.5 m x 0.512°C/m ) + 100.0°C
Tb (Solution) = 4.352 + 100.0°C
Tb (Solution) = 104.352°C
so the boiling point of Mg₃(PO₄)₂ 8.5 m solution = 104.352°C
