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solmaris [256]
3 years ago
15

A solution at 25°C is 1.0 x 10^-5 M H3O+. What is the concentration of OH- in this solution

Chemistry
1 answer:
Kobotan [32]3 years ago
7 0

Answer:
1.0 × 10–9 M OH–

Explanation:
pH = -Log[H+]
pOH = -Log[OH-]
But;
pH + pOH = 14
Therefore;
[H+] + [OH-] = 1.0 × 10^-14 M
Therefore;
[OH-] = 1.0 × 10^-14 M - (1.0 × 10^–5 M)
= 1.0 × 10^-9 M OH–
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Colin is writing a story in which a spy damages a fusion nuclear reactor, the reaction runs out of control, and the reactor expl
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The phenomenon that releases uncontrolled amounts of energy very fast is fission, and that is what is used in bombs.
The last option is correct.
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3 years ago
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The two algae seen here are classified as green algae. That means they make their own food by photosynthesis. The algae on the l
ohaa [14]

Answer:

C

Explanation:

Unicellular algae can't do photosynthesis because it is a single cell organisms, therefore the answer is C

6 0
3 years ago
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If you start with 6 moles of N2 and 6 moles of H2 (meaning you won't have enough of 1 of the ingredients), how many moles of NH3
Ivahew [28]

Answer:

4molNH_3

Explanation:

Hello there!

In this case, according to the given information it will be firstly necessary to set up the chemical equation taking place:

N_2+3H_3\rightarrow 2NH_3

We infer we need to calculate the moles of NH3 by using both of the moles of N2 and H2 at the beginning, in order to identify the limiting reactant:

n_{NH_3}=6molN_2*\frac{2molNH_3}{1molN_2}=12molNH_3\\\\ n_{NH_3}=6molH_2*\frac{2molNH_3}{3molH_2}=4molNH_3\\

Thus, since hydrogen yields the fewest moles of ammonia, we conclude that we are just able to yield 4 moles of NH3.

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8 0
3 years ago
19
N76 [4]

Answer:

no. of moles of PbI₂ = 14 moles

Explanation:

Data Given:

no. of moles of HI = 28 moles

no. of moles of Pbl₂ = ?

Reaction Given

        Pb + 2HI -------→ Pbl₂ + H₂

Solution:

To solve this problem we have to look at the reaction

Reaction:

                  Pb  +  2HI   -------→  Pbl₂  +  H₂

                            2 mol          1 mol

So if we look at the reaction 2 mole of hydrogen iodide (HI) gives 1 moles of Lead iodide (PbI₂), then how many moles of Lead iodide (PbI₂) will be produced by 28 moles of hydrogen iodide (HI)

For this apply unity formula

                      2 mole of HI  ≅  1 moles of PbI₂

                      28 mole of HI  ≅  X moles of PbI₂

By Doing cross multiplication

                      moles of PbI₂ = 28 moles x 1 moles / 2 mole

                      moles of PbI₂ = 14 mole

28 mole of HI will produce 14 moles of PbI₂

4 0
3 years ago
In a laboratory experiment you are given 46.1 grams of MgSO4 and asked to make a 1.20 M solution. The volume of your solution wi
Darina [25.2K]

Answer:

0.319 L

Explanation:

M(MgSO4) = 120 g/mol

46.1 g * 1 mol/120 g = 0.384 mol MgSO4

0.384 mol * 1 L/1.20 mol = 0.319 L

5 0
3 years ago
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