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Liula [17]
2 years ago
8

Lab Report: Paper Chromatography

Chemistry
1 answer:
vampirchik [111]2 years ago
7 0

Answer:

the answer is "I HOPE IT HELPS"

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Which activity demonstrates chemical weathering?
valina [46]

The answer is A, to my brain power ig B)

6 0
3 years ago
Is more chemical weathering likely to occur in New York location or in the Illinois location? Explain your answer. HELP NEEDED (
Naily [24]
The pH of pure water is 7.0, which is neutral.

For a pH below this, the water is acidic. Substances that are acidic are often corrosive and thus could cause weather damaging. The pH of the precipitation in NY is below that of the precipitation in IL, and NY receives more precipitation, so for both of those reasons, it is likely to have more chemical weathering.
3 0
3 years ago
Zinc + Hydrochloric acid — Zinc chloride + Hydrogen [Zn +2 HCI → ZnCl2 + H2] is an
Jet001 [13]
Double replacement i believe
6 0
3 years ago
What is the hydronium ion concentration of a 0.100 M acetic acid solution with Ka = 1.8 × 10-5? The equation for the dissociatio
evablogger [386]

Answer:

1.3×10⁻³ M

Explanation:

Hello,

In this case, given the dissociation reaction of acetic acid:

CH_3CO_2H(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + CH_3CO_2^-(aq)

We can write the law of mass action for it:

Ka=\frac{[H_3O^+][CH_3CO_2^-]}{[CH_3CO_2H]}

Of course, excluding the water as heterogeneous substances are not included. Then, in terms of the change x due to the dissociation extent, we are able to rewrite it as shown below:

1.8x10^{-5}=\frac{x*x}{0.100-x}

Thus, via the quadratic equation or solve, we obtain the following solutions:

x_1=-0.00135M\\x_2=0.00133M

Obviously, the solution is 0.00133M which match with the hydronium concentration, thus, answer is: 1.3×10⁻³ M in scientific notation.

Regards.

8 0
3 years ago
Read 2 more answers
Hydrogen 3 has a half life of 12.32 years a sample of h-3 weighing 3.02 grams is left for 15.0 years what will the final weight
yawa3891 [41]

Answer:

The final mass of sample is 1.3 g.

Explanation:

Given data:

Half life of H-3 = 12.32 years

Amount left for 15.0 years = 3.02 g

Final amount = ?

Solution:

First all we will calculate the decay constant.

t₁/₂ = ln² /k

t₁/₂ =12.32 years

12.32 y =  ln² /k

k = ln²/12.32 y

k = 0.05626 y⁻¹

Now we will find the original amount:

ln (A°/A) = Kt

ln (3.02 g/ A) = 0.05626 y⁻¹ × 15.0 y

ln (3.02 g/ A) = 0.8439

3.02 g/ A = e⁰°⁸⁴³⁹

3.02 g/ A = 2.33

A = 3.02 g/ 2.33

A = 1.3 g

The final mass of sample is 1.3 g.

8 0
3 years ago
Read 2 more answers
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