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3 years ago

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Compare and contrast replication and repeated trials

1 answer:

irina [24]3 years ago

8 0

<span>Researching is composed of various explorations in utilizing the scientific method in the process. It also, answers any scientific inquiry with the use of scientific method. Research enables and promotes the scientific community at a larger scale. Contributing and collaborating knowledge all-over the people and persons in science. Researches play a big role in everyone’s academic identity because </span>
<span><span>1. </span>It actively makes the individual scientific in approach to things of curiosity and makes him/her use the knowledge to study and produce results which</span> <span><span>
2. </span>The scientific society will benefit by this particular study and can work together to better explore and discover. </span><span> </span>

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D is the answee your welcome

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3 years ago

What are tissues made up of?

Aleks [24]

Answer should be:

Cells

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2 years ago

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Which of the following are examples of useful hydrocarbons?

kari74 [83]

Answer:

the most useful hydrocarbon is fuel.

Explanation:

some examples like: natural gases,coal,diesel fuel, kerosene etc.

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1 year ago

A weak monoprotic acid is titrated with 0.100 MNaOH. It requires 50.0 mL of the NaOH solution to reach the equivalence point. Af

larisa86 [58]

Explanation:

The given reaction is as follows.

$HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)$

Hence, number of moles of NaOH are as follows.

n = $0.05 L \times 0.1 M$

= 0.005 mol

After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.

n = $0.025 L \times 0.1 M$

= 0.0025 mol

According to ICE table,

$HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)$

Initial: 0.005 mol 0.0025 mol 0 0

Change: -0.0025 mol -0.0025 mol +0.0025 mol

Equibm: 0.0025 mol 0 0.0025 mol

Hence, concentrations of HA and NaA are calculated as follows.

[HA] = $\frac{0.0025 mol}{V}$

[NaA] = $\frac{0.0025 mol}{V}$

$[A^{-}] = [NaA] = \frac{0.0025 mol}{V}$

Now, we will calculate the $pK_{a}$ value as follows.

pH = $pK_{a} + log \frac{A^{-}}{HA}$

$pK_{a} = pH - log \frac{[A^{-}]}{[HA]}$

= $3.42 - log \frac{\frac{0.0025 mol}{V}}{\frac{0.0025}{V}}$

= 3.42

Thus, we can conclude that $pK_{a}$ of the weak acid is 3.42.

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3 years ago

Please help im rlly stuck

atroni [7]

Answer:

A?

Explanation:

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2 years ago

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