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3 years ago

What is the pOH of a solution with [OH-] = 1.4 x 10-13?

Chemistry

1 answer:

lukranit [14]3 years ago

4 0

Answer:

C. 12.85

Explanation:

just did

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How many molecules are in 3.5 moles of H2O A. 2.107x10^24 B. 3.87x10^24 C. 21.07 B. 1.72x10^23

Pani-rosa [81]

Answer:

A. 2.107x10^24

Explanation:

1 mole is defined as the particles containing in 12g of 12-C. That is 6.022x10²³ particles.

When you say 1 mole of atoms means 6.022x10²³ atoms, 1 mole of molecules means 6.022x10²³ molecules, etc.

Now, 3.5 moles of H₂O are:

3.5mol * (6.022x10²³ molecules / 1mol) = 2.107x10²⁴ molecules of H₂O.

Thus, right answer is:

8 0

3 years ago

1. Shankar and Sameer performed an experiment to differentiate primary, secondary and tertiary amines in a laboratory.Shankar co

lesya [120]

Answer: 2) Chloroform & Caustic potash

Explanation:

The carbylamine reaction is a kind of chemical test which is done to detect primary amines in an unknown solution. It cannot detect secondary and tertiary amines.

The reaction involves the heating with up of the unknown solution with alcoholic potassium hydroxide or caustic potash and the chloroform.

In the presence of primary amine, the production of isocyanide results.

3 0

3 years ago

What is the correct name for the molecular compound n2s4?

blsea [12.9K]

Dinitrogen tetrasulfide

3 0

3 years ago

When the concentration of hydronium ions equals the concentration of hydroxide ions, it is called the _____. balence?

GrogVix [38]

The answer is Equivalence point. When the condition of [H+] = [OH-<span>] is reached, this state is called the </span><span>equivalence point.</span>

6 0

3 years ago

When 0.100 mol of carbon is burned in a closed vessel with8.00

antoniya [11.8K]

Answer : The mass of carbon monoxide form can be 2.8 grams.

Solution : Given,

Moles of C = 0.100 mole

Mass of $O_2$ = 8.00 g

Molar mass of $O_2$ = 32 g/mole

Molar mass of CO = 28 g/mole

First we have to calculate the moles of $O_2$ .

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{8g}{32g/mole}=0.25moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2C+O_2\rightarrow 2CO$

From the balanced reaction we conclude that

As, 2 mole of $C$ react with 1 mole of $O_2$

So, 0.1 moles of $C$ react with $\frac{0.1}{2}=0.05$ moles of $O_2$

From this we conclude that, $O_2$ is an excess reagent because the given moles are greater than the required moles and $C$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CO$

From the reaction, we conclude that

As, 2 mole of $C$ react to give 2 mole of $CO$

So, 0.1 moles of $C$ react to give 0.1 moles of $CO$

Now we have to calculate the mass of $CO$

$\text{ Mass of }CO=\text{ Moles of }CO\times \text{ Molar mass of }CO$

$\text{ Mass of }CO=(0.1moles)\times (28g/mole)=2.8g$

Therefore, the mass of carbon monoxide form can be 2.8 grams.

5 0

3 years ago