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lilavasa [31]
3 years ago
15

you have a solution of water and salt. if the total solution weighs 50 grams and the salt weighs 4 grams. what is the percent co

ncentration of salt​
Chemistry
1 answer:
Natali5045456 [20]3 years ago
8 0

Answer:

92%

Explanation:

I believe. Hope this was helpful.

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State briefly the similarities and differences between the theories of Arrhenius and Bronsted-Lowry
ludmilkaskok [199]
My answer to the question above is not the best example but I hope it will help you. <span>The Arrhenius model says that acids always contain H+ and that bases always contain OH-. </span>
<span>The Bronsted-Lowry model thinks of acids as being proton donors and proton acceptors, so bases no longer need to contain OH-, and acids donate a proton to water forming H3O+. </span>
<span>Lewis acids are electron pair acceptors, and Lewis bases are electron pair donors. For instance, H+ + OH- => H20. H+ has no electrons, so when it bonds to the Oxygen, it gains an electron pair. OH- "loses" an electron pair.</span>
4 0
3 years ago
A ______ is a process in which the physical and chemical properties of the original substances change as new substances are form
Jlenok [28]

Answer:

chemical change

Explanation:

In chemical change , considerable amount of heat is used , hence a change in substance formed

7 0
3 years ago
Read 2 more answers
Any molecules can have this force between them?.
RoseWind [281]

Answer:

Intermolecular bonds are found in all of them. They can come in different versions.

5 0
3 years ago
The half-life of radium-226 is 1590 years. if a sample contains 100 mg, how many mg will remain after 1000 years?
AlladinOne [14]

Answer:

64.52 mg.

Explanation:

The following data were obtained from the question:

Half life (t½) = 1590 years

Initial amount (N₀) = 100 mg

Time (t) = 1000 years.

Final amount (N) =.?

Next, we shall determine the rate constant (K).

This is illustrated below:

Half life (t½) = 1590 years

Rate/decay constant (K) =?

K = 0.693 / t½

K = 0.693/1590

K = 4.36×10¯⁴ / year.

Finally, we shall determine the amount that will remain after 1000 years as follow:

Half life (t½) = 1590 years

Initial amount (N₀) = 100 mg

Time (t) = 1000 years.

Rate constant = 4.36×10¯⁴ / year.

Final amount (N) =.?

Log (N₀/N) = kt/2.3

Log (100/N) = 4.36×10¯⁴ × 1000/2.3

Log (100/N) = 0.436/2.3

Log (100/N) = 0.1896

Take the antilog

100/N = antilog (0.1896)

100/N = 1.55

Cross multiply

N x 1.55 = 100

Divide both side by 1.55

N = 100/1.55

N = 64.52 mg

Therefore, the amount that remained after 1000 years is 64.52 mg

7 0
4 years ago
I need help solving this!
zmey [24]

Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, CH_{4}.

Explanation:

Given: Mass of methane = 146.6 g

As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.

Moles = \frac{mass}{molar mass}\\= \frac{146.6 g}{16.04 g/mol}\\= 9.14 mol

The given reaction equation is as follows.

C + 2H_{2} \rightarrow CH_{4}

This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.

Moles of H_{2} = \frac{9.14}{2}\\= 4.57 mol

Thus, we can conclude that moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, CH_{4}.

5 0
3 years ago
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