Given:
114.32g of water
14.85oC is raised
to 18.00oC
Required:
Change in heat
energy
Solution:
This can be solved
through the equation H = mCpT
where H is the heat, m is the mass, Cp is the specific heat and T is the change in temperature.
The specific heat
of the water is 4.18 J/g-K
Plugging in the
values into the equation
H = mCpT
H = (114.32g) (4.18
J/g °C) (18 – 14.85)
H = 1,505.3 J
(11.21 g BaSO4) / (233.4 g/mol BaSO4) = 0.0480 mol BaSO4 and original barium salt
<span>(10.0 g) / (0.0480 mol) = 208.3 g/mol </span>
<span>So it must have been BaCl2.
I think it letter A.
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
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A and D
Density is mass over volume, if mass decreases or volume increases density decreases
Answer:
Explanation:
In one mole of liquid HF , no of molecules is 6.02 x 10²³.
Each molecule forms two hydrogen bonds on an average so no of hydrogen bonds per mol ih HF is
2 x 6.02 x 10²³.
= 12.04 x 10²³ .
In water no of hydrogen bonds per molecule of water is 4 . So in one mole of water no of hydrogen bonds is
4 x 6.02 x 10²³.
= 24.08 x 10²³ .
Answer:
Whether barium chloride solution was pure
Explanation:
We may answer whether barium chloride was pure. The sequence of this experiment might be depicted by the following balanced chemical equations:
Having a total sample of 10.0 grams, we would firstly find the mass percentage of barium in barium chloride:
This means in 10.0 g, we have a total of:
of barium cations.
The precipitate is then formed and we measure its mass. Having its mass determined, we'll firstly find the percentage of barium in barium sulfate using the same approach:
Multiplying the mass we obtained by the fraction of barium will yield mass of barium in barium sulfate. Then:
- if this number is equal to 6.595 g, we have a pure sample of barium chloride;
- if this number is lower than 6.595 g, this means we have an impure sample of barium chloride, as we were only able to precipitate a fraction of 6.595 g.
