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Marat540 [252]
3 years ago
5

How many mols are present in the a sample of silver nitrate which has 5.3x10^24 molecules.

Chemistry
1 answer:
Westkost [7]3 years ago
5 0

Answer:  5.3 x 10^24 formula units of silver nitrate is equivalent to 8.8 moles of silver nitrate. Silver nitrate is an ionic compound, therefore, its representative particle is called a "formula unit" instead of molecule. For every mole of a substance, we know that there are 6.022 x 10^23 representative units of that substance. The amount of particles in one mole of substance is called Avogadro's number.

Further Explanation:

We can convert from number of representative particles to moles using the formula:

\boxed {no. \ of \ moles \ = \ ( given \ no. \ of \ particles) \ (\frac{1 \ mole}{\ 6.022 \ x 10^{23} particles})}

For this problem, we can calculate the number of moles by plugging in the given values to the equation above,

no. \ of \ moles \ = (5.3 \ x \ 10^{24} \ formula \ units \ AgNO_{3}) \ (\frac{1 \ mole \ AgNO_{3}}{6.022 \ x 10^{23} \ formula \ units AgNO_{3}}) \\\\\boxed {no. \ of moles \ AgNO_{3} \ = \ 8.8 \ moles}

Learn More

  1. Learn more about representative particles brainly.com/question/8969313
  2. Learn more about Avogadro's number brainly.com/question/229300
  3. Learn more about mole conversions brainly.com/question/1370888

Keywords: moles conversion, Avogadro's number

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sample of atmospheric gas collected at an industrial site is stored in a 250 mL amber glass bottle that has a pressure of 1.02 a
Nady [450]

Answer:- New pressure is 0.942 atm.

Solution:- The volume of the glass bottle would remain constant here and the pressure will change with the temperature.

Pressure is directly proportional to the kelvin temperature. The equation used here is:

P_1T_2=P_2T_1

Where, T_1 and T_2 are initial and final temperatures, P_1 and P_2 are initial and final pressures.

T_1 = 20.3 + 273.15 = 293.45 K

T_2 = -2.0 + 273.15 = 271.15 K

P_1 = 1.02 atm

T_2  = ?

Let's plug in the values in the equation and solve it for final pressure.

1.02atm(271.15K)=P_2(293.45K)

P_2=\frac{1.02atm*271.15K}{293.45K}

P_2 = 0.942 atm

So, the new pressure of the jar is 0.942 atm.


5 0
3 years ago
Why do people want different kinds of light bulbs?
ololo11 [35]

Answer:

LED bulbs fit standard light sockets and are the most energy-efficient option. LEDs have lower wattage than incandescent bulbs but emit the same light output. This allows them to produce the same amount of light but use less energy. LEDs can last over 20 years and don't contain mercury

7 0
2 years ago
A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of
Bad White [126]

Answer:

a) T_{2} = 360.955\,K, P_{2} = 138569.171\,Pa\,(1.386\,bar), b) T_{2} =  347.348\,K, V_{2} = 0.14\,m^{3}

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

\frac{T_{1}}{P_{1}} = \frac{T_{2}}{P_{2}}

Final temperature is cleared from this expression:

Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})

T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}

The number of moles of the ideal gas is:

n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}

n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}

n = 5.541\,mol

The final temperature is:

T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}

T_{2} = 360.955\,K

The final pressure is:

P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}

P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)

P_{2} = 138569.171\,Pa\,(1.386\,bar)

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}

Final temperature is cleared from this expression:

Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})

T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}

T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}

T_{2} =  347.348\,K

The final volume is:

V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}

V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})

V_{2} = 0.14\,m^{3}

4 0
3 years ago
The al alkaline earth metals have two railing electrons how does it form ions 1. Gain two electrons 2. Gain 6 electrons 3. They
S_A_V [24]
The correct response would be 3. The alkaline earth metals would tend to lose its valence electrons, in this case 2 of them at different energy levels, to form the same respective ion, which is +2.
5 0
3 years ago
Modify the given fatty acid so that it represents the 18‑carbon fatty acid designated 18:2(Δ9,12). Draw any double bonds in the
Mekhanik [1.2K]

Answer:

See explanation

Explanation:

In this case, we have to remember the meaning of the nomenclature "18:2Δ9,12".  Where 18 is the <u>number of carbon atom</u>s, 2 is the <u>number of double bonds,</u> and the numbers successive to Δ "delta" the position of the double bonds <u>starting</u> to count from the carboxylic -COOH end of the molecule.

In other words, the main functional group is a <u>carboxylic acid</u>. We have a total of 18 carbons. Additionally, we have 2 double bonds. On carbons 9 and 12.

Lets see figure 1

I hope it helps!

5 0
3 years ago
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