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sergey [27]
3 years ago
10

PLEASE HELP THIS IS DUE IN AN HOUR!!!

Chemistry
1 answer:
Soloha48 [4]3 years ago
7 0

Explanation:

The hydronium ion concentration can be found from the pH by the reverse of the mathematical operation employed to find the pH. [H3O+] = 10-pH or [H3O+] = antilog (- pH) Example: What is the hydronium ion concentration in a solution that has a pH of 8.34? On a calculator, calculate 10-8.34, or "inverse" log ( - 8.34).

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Xelga [282]

Answer:

Top layer is Organic (CH2Cl2 and product)

Explanation:

In a solvent mixture, there are usually two phases, the organic phase and the aqueous phase.

It is usual that the organic phase is almost always less dense than the aqueous phase hence the organic phase tend to remain on top of the aqueous phase.

Hence, the top layer is expected to be the organic CH2Cl2 and product.

8 0
3 years ago
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Explanation:

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2 years ago
Determine the concentration of H+ ions in an aqueous solution where [OH-] =2.25 × 10^-4 M. I have 29 minutes I beg u guys pls hu
mixas84 [53]

but where Is the volume in order for us to determine the concentration. since we have moles in H+ ions

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6 0
3 years ago
1. A student tried to solve the following problem by selecting the tile as shown. What, if anything, did the student do wrong?
slega [8]

Answer:

B. The student chose the correct tile, but needs to flip the tile to make the units cancel

Explanation:

Based on the reaction:

2AgNO₃(aq) + Cu(s) → 2Ag(s) + Cu(NO₃)₂ (aq)

<em>2 moles of AgNO₃ react per mole of Cu producing 2 moles of Ag and 1 mole of Cu(NO₃)₂</em>

Thus, if you want to produce 6.75moles of Cu(NO₃)₂ you need:

6.75moles of Cu(NO_3)_2  \frac{2 mol AgNO_3 }{1 mol Cu(NO_3)_2} = 13.50 moles of AgNO₃ are needed

Thus, if you analize the tile shown by the student:

<em>B. The student chose the correct tile, but needs to flip the tile to make the units cancel</em>

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3 years ago
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Truly loving someone means caring for them in the ways that they need to be cared for, with no strings attached. (That's why they call it unconditional love.)

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