Answer:
The answer to your question is: letter D. 1.33 L
Explanation:
Data
V1 = 50 ml
C1 = 19.3
To solve this problem use the formula C₁V₁ = C₂V₂
C2 = C1V1 / V2
C = concentration
V = volume
a) 1.15 L
C2 = (19.3)(50) / 1150
C2 = 0.84 M
b) No right answer
c) V2= 0.80 L
C2 = (19.3)(50) / 800
C2 = 1.2 M
d) V2 = 1.33 L
C2 = (19.3)(50) / 1330
C2 = 0.72 M
e) V2 = 350 ml
C2 = (19.3)(50) / 350
C2 = 2.75 M
Answer:
Sublimation is the transition of a substance directly from the solid to the gas state, without passing through the liquid state. ... The reverse process of sublimation is deposition or desublimation, in which a substance passes directly from a gas to a solid phase.
I think the Ksp for Calcium Carbonate is around 5×10⁻⁹
(I don't know if this is the Ksp value that you use because I read somewhere that this value can vary. You should probably check with your teacher with what Ksp value they want you to use)
the equation for the dissociation CaCO₃ in water is CaCO₃(s)⇄Ca²⁺(aq)+CO₃²⁻(aq) which means that the concentration of Ca²⁺ is equal to the concentration of CO₃²⁻ in solution. For every molecule of CaCO₃ that dissolves, one atom of Ca²⁺ and one molecule of CO₃²⁻ is put into solution which is why the concentrations are equal in solution.
Since Ksp=[Ca²⁺][CO₃²⁻] and we know that [Ca²⁺]=[CO₃²⁻] we can rewrite the equation as Ksp=x² since if you say that [Ca²⁺]=[CO₃²⁻] when you multiply them together you get the concentration squared (I am calling the concentration x for right now).
when solving for x:
5×10⁻⁹=x²
x=0.0000707
Therefore [Ca²⁺]=[CO₃²⁻]=0.0000707mol/L which also shows how much calcium carbonate is dissolved per liter of water since the amount of Ca²⁺ and CO₃²⁻ in solution came from the calcium in a 1 to 1 molar ratio as shown in the equation (the value we found for x is the molar solubility of calcium carbonate).
Using the fact that the molar mass of calcium carbonate is 100.09g/mol you can use dimensional analysis as fallows:
(0.0000707mol/L)(100.09g/mol)=0.007077g/L
That means that there is 0.007077g of Calcium carbonate that can precipitate out of 1L of water.
since the question is asking for how much water needs to be evaporated to precipitate 100mg (0.1g) of Calcium you have to do the fallowing calculation:
(0.1g)/(0.007077g/L)=14.13L of water.
14.13L of water needs to evaporate in order to precipitate out 100mg of calcium carbonate
These types of questions can get long and confusing so I bolded parts that were important to try to guide you through it more easily.
I hope this helps. Let me know if anything is unclear.
Answer:
Q14: 17,140 g = 17.14 kg.
Q16: 504 J.
Explanation:
<u><em>Q14:</em></u>
- To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat absorbed by ice (Q = 3600 x 10³ J).
m is the mass of the ice (m = ??? g).
c is the specific heat of the ice (c of ice = 2.1 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 100.0°C - 0.0°C = 100.0°C).
∵ Q = m.c.ΔT
∴ (3600 x 10³ J) = m.(2.1 J/g.°C).(100.0°C)
∴ m = (3600 x 10³ J)/(2.1 J/g.°C).(100.0°C) = 17,140 g = 17.14 kg.
<u><em>Q16:</em></u>
- To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat absorbed by ice (Q = ??? J).
m is the mass of the ice (m = 12.0 g).
c is the specific heat of the ice (c of ice = 2.1 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 0.0°C - (-20.0°C) = 20.0°C).
∴ Q = m.c.ΔT = (12.0 g)(2.1 J/g.°C)(20.0°C) = 504 J.
Answer:
b
Explanation:
electrons lie in the orbits of the atom. they are negatively charged and they are very small in size. they have very less mass than proton. ele tron has 1/1840 mass of proton.
