Answer:
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to solve for the molarity of the KOH solution by knowing this base react in a 1:1 mole ratio with nitric acid, HNO3; thus, we can write the following equation, as their moles are the same at the endpoint:
Which in terms of molarities and volumes is:
Thus, we solve for the molarity of the base (KOH) to obtain:
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Hello ur answer will be D, hope this helps :)
Answer:
66 g of CO₂
Solution:
The Balance Chemical Reaction is as follow,
C₂H₂ + 5/2 O₂ → 2 CO₂ + H₂O
Or,
2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O ------- (1)
Step 1: Find out the limiting reagent as;
According to Equation 1,
56.1 g (2 mole) C₂H₂ reacts with = 160 g (5 moles) of O₂
So,
125 g of C₂H₂ will react with = X g of O₂
Solving for X,
X = (125 g × 160 g) ÷ 56.1 g
X = 356.5 g of O₂
It means for total combustion of Ethylene we require 356.5 g of O₂, but we are only provided with 60.0 g of O₂. Therefore, O₂ is the limiting reagent and will control the yield.
Step 2: Calculate Amount of CO₂ produced as;
According to Equation 1,
160 g (5 mole) O₂ produces = 176 g (4 moles) of CO₂
So,
60.0 g of O₂ will produce = X g of CO₂
Solving for X,
X = (60.0 g × 176 g) ÷ 160 g
X = 66 g of CO₂
Answer:
The heat required to change 25.0 g of water from solid ice to liquid water at 0°C is 8350 J
Explanation:
The parameters given are
The temperature of the solid water = 0°C
The heat of fusion, = 334 J/g
The heat of vaporization, = 2260 J/g
Mass of the solid water = 25.0 g
We note that the heat required to change a solid to a liquid is the heat of fusion, from which we have the formula for heat fusion is given as follows;
ΔH = m ×
Therefore, we have;
ΔH = 25 g × 334 J/g = 8350 J
Which gives the heat required to change 25.0 g of water from solid ice to liquid water at 0°C as 8350 J.
Increase: volume, pressure, kinetic energy, gas particle collisions
Stays the same: number of moles of gas
