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Dafna1 [17]
2 years ago
15

What are the number of atoms in each Chemical Formula

Chemistry
1 answer:
podryga [215]2 years ago
6 0
You can search that up online it’s not that hard but good luck !!
You might be interested in
Please help Please help
alexdok [17]
Follow Avogadro’s Number
1 mole = 6.02 x 10^23
So we can do it
4.77x10^25/6.02x10^23 = 79.2 mole
5 0
1 year ago
The acid-dissociation constants of HC3H5O3 and CH3NH3+ are given in the table below. Which of the following mixtures is a buffer
sergey [27]

Answer:

A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH

Explanation:

The pH of a buffer solution is calculated using following relation

pH=pKa+log(\frac{salt}{acid} )

Thus the pH of buffer solution will be near to the pKa of the acid used in making the buffer solution.

The pKa value of HC₃H₅O₃ acid is more closer to required pH = 4 than CH₃NH₃⁺ acid.

pKa = -log [Ka]

For HC₃H₅O₃

pKa = 3.1

For CH₃NH₃⁺

pKa = 10.64

pKb = 14-10.64 = 3.36 [Thus the pKb of this acid is also near to required pH value)

A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH

Half of the acid will get neutralized by the given base and thus will result in equal concentration of both the weak acid and the salt making the pH just equal to the pKa value.

8 0
2 years ago
1. What is the probability of rolling a five on one die three times in a row?
Bad White [126]
1/6 x 1/6 x 1/6= solve the rest
7 0
3 years ago
Calculate the freezing point and boiling point of a solution containing 8.15 g of ethylene glycol (C2H6O2) in 96.3 mL of ethanol
pishuonlain [190]

<u>Answer:</u> The freezing point of solution is -117.54°C and the boiling point of solution is 80.48°C

<u>Explanation:</u>

To calculate the mass of ethanol, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of ethanol = 0.789 g/mL

Volume of ethanol = 96.3 mL

Putting values in above equation, we get:

0.789g/mL=\frac{\text{Mass of ethanol}}{96.3mL}\\\\\text{Mass of ethanol}=(0.789g/mL\times 96.3mL)=75.98g

  • <u>Calculating the freezing point:</u>

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = -114.1 °C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point elevation constant = 1.99°C/m

m_{solute} = Given mass of solute (ethylene glycol) = 8.15 g

M_{solute} = Molar mass of solute (ethylene glycol) = 62 g/mol

W_{solvent} = Mass of solvent (ethanol) = 75.98 g

Putting values in above equation, we get:

-114.1-\text{Freezing point of solution}=1\times 1.99^oC/m\times \frac{8.15\times 1000}{62g/mol\times 75.98}\\\\\text{Freezing point of solution}=-117.54^oC

Hence, the freezing point of solution is -117.54°C

  • <u>Calculating the boiling point:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

Or,

\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

where,

Boiling point of pure solution = 78.4°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_b = molal boiling point elevation constant = 1.20°C/m.g

m_{solute} = Given mass of solute (ethylene glycol) = 8.15 g

M_{solute} = Molar mass of solute (ethylene glycol) = 62  g/mol

W_{solvent} = Mass of solvent (ethanol) = 75.98 g

Putting values in above equation, we get:

\text{Boiling point of solution}-78.4=1\times 1.20^oC/m\times \frac{8.15\times 1000}{62\times 75.98}\\\\\text{Boiling point of solution}=80.48^oC

Hence, the boiling point of solution is 80.48°C

3 0
3 years ago
What volume will be occupied by .756 mole of gas at 109 kPa and 30.5 degrees C
olga2289 [7]

Answer: it would be 0.026 moles

Explanation: PV=nRT, P is the pressure of gas, V is the volume it occupies n is the number of moles of gas present in the sample, R is the universal gas constant which is equal to 0.0821 atm L/mol K and T is the absolute temperature of the gas

8 0
3 years ago
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