Answer:
0.297 °C
Step-by-step explanation:
The formula for the <em>freezing point depression </em>ΔT_f is
ΔT_f = iK_f·b
i is the van’t Hoff factor: the number of moles of particles you get from a solute.
For glucose,
glucose(s) ⟶ glucose(aq)
1 mole glucose ⟶ 1 mol particles i = 1
Data:
Mass of glucose = 10.20 g
Mass of water = 355 g
ΔT_f = 1.86 °C·kg·mol⁻¹
Calculations:
(a) <em>Moles of glucose
</em>
n = 10.20 g × (1 mol/180.16 g)
= 0.056 62 mol
(b) <em>Kilograms of water
</em>
m = 355 g × (1 kg/1000 g)
= 0.355 kg
(c) <em>Molal concentration
</em>
b = moles of solute/kilograms of solvent
= 0.056 62 mol/0.355 kg
= 0.1595 mol·kg⁻¹
(d) <em>Freezing point depression
</em>
ΔT_f = 1 × 1.86 × 0.1595
= 0.297 °C
So to solve this you need to know Charles’s law which is: V1/T1=V2/T2. Where T1 and V1 is the initial volume and Temperature and V2 and T2 is the temperature and volume afterwards. So first plug in the numbers you are given. V1= 1.55L T1= 32C° V2= 755mL T2=?. Since your volumes are two different units you change 755mL to be in L so that would be 0.755 L. And since your temp isn’t in Kelvin you do 273+32= 305K°. You then would rearrange your equation to solve for T2 which is V2T1/V1. Then you plug in your numbers (0.755L)(305K)/1.55L. Then you solve and would be 148.5645161 —> 1.49 x 10^2 K
Answer:
About 110 g.
Your tool of choice here will be the solubility graph for potassium nitrate, KNO3, in water.
Answer:
+4
Explanation:
In PbO2, oxygen exhibits an oxidation number of -2 (since it's not a peroxide or superoxide):
Let the oxidation number of Pb be x. Then, for the compound to be neutral, the oxidation numbers of all atoms should add up to zero.
⇒ x + (−2) + (−2) = 0
x = +4
So the oxidation no. of Pb is +4.
I hope this helps.
Answer : The metal used was iron (the specific heat capacity is 
).
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
where,
= specific heat of unknown metal = ?
= specific heat of water = 
= mass of unknown metal = 150 g
= mass of water = 200 g
= final temperature of water = 
= initial temperature of unknown metal = 
= initial temperature of water = 
Now put all the given values in the above formula, we get
Form the value of specific heat of unknown metal, we conclude that the metal used in this was iron (Fe).
Therefore, the metal used was iron (the specific heat capacity is ).
