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Temka [501]
2 years ago
7

At which point does condensation occur in this water cycle diagram?

Chemistry
2 answers:
Nata [24]2 years ago
7 0

Answer:

it would be b

Explanation:

b would be condensation

Ymorist [56]2 years ago
4 0

Answer: B

Explanation:

A is evaporation. B is condensation. C is precipitation.

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Nady [450]

Answer:

taxonomy

Explanation:

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3 years ago
Do viruses respond to stimuli
sertanlavr [38]
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7 0
2 years ago
Read 2 more answers
A patient needs to be given exactly 1000 ml of a 5.0% (w/v) intravenous glucose solution. the stock solution is 55% (w/v). how m
LiRa [457]
The patient needs 1000 ml of 5% (w/v) glucose solution
i.e  1000 ml x 5 g/ 100 ml 
where the stock solution is 55% (w/v) = 55 g / 100 ml  
So, 1000 ml x 5 g / 100 ml = V (ml) x 55 g / 100 ml 
V = 1000 x (5 / 100) / (55 / 100) = 5000 / 55 = 90.9 ml 
∴ the patient needs 90.9 ml of 55% (w/v) glucose solution

6 0
3 years ago
What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb
Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

(\text{CH}_3)_3\text{N} in this question acts as a weak base. As seen in the equation in the question, (\text{CH}_3)_3\text{N} produces \text{OH}^{-} rather than \text{H}^{+} when it dissolves in water. The concentration of \text{OH}^{-} will likely be more useful than that of \text{H}^{+} for the calculations here.

Finding the value of [\text{OH}^{-}] from pH:

Assume that \text{pK}_w = 14,

\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}.

[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}.

Solve for [(\text{CH}_3)_3\text{N}]_\text{initial}:

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}

Note that water isn't part of this expression.

The value of Kb is quite small. The change in (\text{CH}_3)_3\text{N} is nearly negligible once it dissolves. In other words,

[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}.

Also, for each mole of \text{OH}^{-} produced, one mole of (\text{CH}_3)_3\text{NH}^{+} was also produced. The solution started with a small amount of either species. As a result,

[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}.

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3},

[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b},

[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}.

8 0
3 years ago
What is the molar concentration of hydronium ions in a sample of a soft drink that has a ph of 4?
hammer [34]

[H^{+}] value of soft drink is 0.0001mol/l when given pH is 4.

Given:

pH = 4

Needs to find: [H^{+}]

Formula to find: pH=−log_{10} [H^{+}]

We can put the values of pH in above formula as,

pH=−log_{10} [H^{+}]

4 =−log_{10} [H^{+}]

To calculate hydonium ion concentration, formula is as follows;

[H^{+}] = 10^{-pH}

[H^{+}] = 10^{-4} = 0.0001mol/l

[H^{+}] value of soft drink is 0.0001mol/l

Learn more about pH here:

brainly.com/question/2288405

#SPJ4

6 0
1 year ago
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