There are several various of expressing concentration. For instance, mass percent, volume percent, Molarity, Normality, Molality, etc.
In present case, weight of solute and solvent are given, so it will be convenient to express concentration in terms of mass percent.
Given: weight of solute (Ca2+) = 8500 g
weight of solvent (water) = 490 g.
Therefore, mass of solution = 8500 + 490 = 8990 g
Now, mass percent =
=
= 94.55 %
Answer: Concentration of calcium ions is in this solution is 94.55 % (w/W)
Answer:
Carbon dixoide and water combine to form glucose and oxgen.
I hope this helped! :D
Explanation:
this process by which green plants and some other organisms use sunlight to synthesize foods from carbon dioxide and water. Photosynthesis in plants generally involves the green pigment chlorophyll and generates oxygen as a byproduct.
Answer: 17.78g
Explanation:
Assume there is no heat exchange with the environment, then the amount of heat taken by the steel rod, Q(s), is equal to the amount of heat lost by the water, Q(w), but with opposite sign.
Q(s) = -Q(w)
Remember, Q = mc(ΔΦ)
Where Q = amount of heat
m = mass of steel
c = specific heat capacity of steel
ΔΦ = Initial temperature T1 - Final temperature T2
Q = mc(T1-T2)
Recall, Q(s) = -Q(w). Then,
m(s)*c(s)*(T1s - T2s) = - m(w)*c(w)*(T1w - T2w)
Substituting each values
Note: m(w) = volume of water*density = 75mL*1g/mL = 75g
m(s)*0.452*(21.5-2) = -75*4.18*(21.5-22)
m(s)*8.814 = 156.75
m(s) = 156.75/8.814
m(s) = 17.78g
Therefore, the mass of steel is 17.78g
Answer:
roar
Explanation:
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Answer:
a. Rate = k×[A]
b. k = 0.213s⁻¹
Explanation:
a. When you are studying the kinetics of a reaction such as:
A + B → Products.
General rate law must be like:
Rate = k×[A]ᵃ[B]ᵇ
You must make experiments change initial concentrations of A and B trying to find k, a and b parameters.
If you see experiments 1 and 3, concentration of A is doubled and the Rate of the reaction is doubled to. That means a = 1
Rate = k×[A]¹[B]ᵇ
In experiment 1 and to the concentration of B change from 1.50M to 2.50M but rate maintains the same. That is only possible if b = 0. (The kinetics of the reaction is indepent to [B]
Rate = k×[A][B]⁰
<h3>Rate = k×[A]</h3>
b. Replacing with values of experiment 1 (You can do the same with experiment 3 obtaining the same) k is:
Rate = k×[A]
0.320M/s = k×[1.50M]
<h3>k = 0.213s⁻¹</h3>