Answer:
44.7 degree (3 s.f)
Explanation:
mass of water = 4.10Kg
Initial temperature = 37.5C
Final temperature = x
Heat = 125 kJ = 125000 J
Specific heat capacity = 4180 J/kg°C
These parameters are given by:
H = MCΔT
ΔT = H / MC
ΔT = 125000 / (4.10 * 4180)
ΔT = 7.294
ΔT = T2 - T1
T2 = ΔT + T1
T2 = 7.294 + 37.5
T2 = 44.794 = 44.7 degree (3 s.f)
Answer:
molecular diagram 3 or 2 is correct answer I hope it helps you
Answer : The volume in mL of the sodium carbonate stock solution is 364 mL.
Explanation :
According to dilution law:
where,
= molarity of aqueous sodium carbonate
= molarity of aqueous sodium carbonate stock solution
= volume of aqueous sodium carbonate
= volume of aqueous sodium carbonate stock solution
Given:
= 1.00 M
= 1.58 M
= 575 mL
= ?
Now put all the given values in the above formula, we get:
Therefore, the volume in mL of the sodium carbonate stock solution is 364 mL.
Answer:
34.8 g
Explanation:
Answer:
We have the masses of two reactants, so this is a limiting reactant problem.
We will need a balanced equation with masses, moles, and molar masses of the compounds involved.
1. Gather all the information in one place with molar masses above the formulas and masses below them.
Mᵣ: 123.90 70.91 208.24
P₄ + 20Cl₂ ⟶ 4PCl₅
Mass/g: 46.0 32.0
2. Calculate the moles of each reactant
3. Calculate the moles of PCl₅ we can obtain from each reactant
From P₄:
The molar ratio is 4 mol PCl₅:4 mol P₄
From Cl₂:
The molar ratio is 4 mol PCl₅:20 mol Cl₂
4. Identify the limiting and excess reactants
The limiting reactant is chlorine, because it gives the smaller amount of PCl₅.
The excess reactant is phosphorus.
5. Mass of excess reactant
(a) Moles of P₄ reacted
The molar ratio is 1 mol P₄:20 mol Cl₂
(b) Mass of P₄ reacted
(c) Mass of P₄ remaining
Mass remaining = original mass – mass reacted = (46.0 - 11.18) g = 34.8 g P₄
Answer:
Atoms of the reactant(s) must equal the atoms of the product(s).
Explanation:
According to the Law of Conservation, all atoms of the reactant(s) must equal the atoms of the product(s).
As a result, we need to balance chemical equations. We do this by adding in coefficients to the reactants and/or products. The compound(s) itself/themselves DOES NOT CHANGE.