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3 years ago

The pH of a solution can be calculated using which formula?

Chemistry

1 answer:

VikaD [51]3 years ago

5 0

Answer:

The formula for calculating pH is pH=−log[H_3O+ ]

pH is the negative logarithm (to base 10) of hydronium ion concentration

The pH Formula can also be expressed as

PH= - log[H+ ]

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If a 67.3G rock is dissolved in 2.00L of acid, what is the molar concentration of gold in the acid solution

brilliants [131]

Answer:

[Au] = 0.171 M

Explanation:

For this question, we assume the rock is 100 % gold.

First of all, we determine the moles of gold

67.3 g . 1mol/ 196.97g = 0.342 moles

Molar concentration is defined as the moles of solute, contained in 1L of solution.

Our solution volume is 2L.

M = 0.342 mol / 2L = 0.171

Molar concentration, also called molarity of solution is the most typical unit of concentration.

6 0

2 years ago

write the symbols for the ions that form when potassium and iodine react to form the ionic compound potassium iodide.

Bingel [31]

I believe it isK+ + I-=KI

3 0

3 years ago

Read 2 more answers

Examples:

Lostsunrise [7]

Uhhhhhhhhhhhhhh wut, can you rewrite it?

6 0

3 years ago

What is the energy per mole of photons, in kJ/mol, emitted by an argon ion laser which produces a green beam with a frequency of

Serga [27]

Answer:

Energy per mole of photons = 2.31 × 10^2 KJ/mol

Explanation:

Energy, E = hf;

Where h is Planck's constant = 6.63 ×10^-34, and f is frequency of the photons.

E = 6.63 × 10^-34 × 5.8 × 10^14

E = 3.84 × 10^-22 KiloJoules

I mole of photons contains Avogadro's number of particles, 6.02 × 10^23

Therefore, the energy per mile of photon is 3.84 × 10^-22 KJ × 6.02 × 10^23

Energy per mile of photon = 2.31 × 10^2 KJ/mol

5 0

3 years ago

Zinc metal and aqueous silver nitrate react to give Zn(NO3)2(aq) plus silver metal. When 5.00 g of Zn(s) and solution containing

Fittoniya [83]

Answer : The percent yield is, 83.51 %

Solution : Given,

Mass of Zn = 5.00 g

Mass of $AgNO_3$ = 25.00 g

Molar mass of Zn = 65.38 g/mole

Molar mass of $AgNO_3$ = 168.97 g/mole

Molar mass of Ag = 107.87 g/mole

First we have to calculate the moles of Zn and $AgNO_3$ .

$\text{ Moles of }Zn=\frac{\text{ Mass of }Zn}{\text{ Molar mass of }Zn}=\frac{5.00g}{65.38g/mole}=0.0765moles$

$\text{ Moles of }AgNO_3=\frac{\text{ Mass of }AgNO_3}{\text{ Molar mass of }AgNO_3}=\frac{25.00g}{168.97g/mole}=0.1479moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag$

From the balanced reaction we conclude that

As, 2 mole of $AgNO_3$ react with 1 mole of $Zn$

So, 0.1479 moles of $AgNO_3$ react with $\frac{0.1479}{2}=0.07395$ moles of $Zn$

From this we conclude that, $Zn$ is an excess reagent because the given moles are greater than the required moles and $AgNO_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Ag$

From the reaction, we conclude that

As, 2 mole of $AgNO_3$ react to give 2 mole of $Ag$

So, 0.1479 moles of $AgNO_3$ react to give 0.1479 moles of $Ag$

Now we have to calculate the mass of $Ag$

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

$\text{ Mass of }Ag=(0.1479moles)\times (107.87g/mole)=15.95g$

Theoretical yield of $Ag$ = 15.95 g

Experimental yield of $Ag$ = 13.32 g

Now we have to calculate the percent yield.

$\% \text{ yield}=\frac{\text{ Experimental yield of }Ag}{\text{ Theretical yield of }Ag}\times 100$

$\% \text{ yield}=\frac{13.32g}{15.95g}\times 100=83.51\%$

Therefore, the percent yield is, 83.51 %

7 0

3 years ago