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labwork [276]
3 years ago
11

How many moles are in 264 g of gold, Au?1 point​

Chemistry
2 answers:
lbvjy [14]3 years ago
8 0

Answer: n = 1.34 moles Au

Explanation: For 1 mole of Au it is equal to its molar mass.

264 g Au x 1 mole Au / 197 g Au

= 1.34 moles Au

Cancel out the units of mass which is g and the remaining unit is in moles.

Brut [27]3 years ago
5 0

Answer: 1/2

Explanation:

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Katen [24]

The following statement is True.

8 0
3 years ago
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Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h
bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

We know that,

The relation between the C_p\text{ and }C_v for an ideal gas are :

C_p-C_v=R

As we are given :

C_p=28.253J/K.mole

28.253J/K.mole-C_v=8.314J/K.mole

C_v=19.939J/K.mole

Now we have to calculate the entropy change of the gas.

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. (w=-pdV)

(C) Heat during the process will be,

q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

7 0
3 years ago
If the molecule could move upward without colliding with other molecules, then how high would it go before coming to rest? Give
tankabanditka [31]

The maximum height at which nitrogen molecule will go before coming to rest is 14 kilometers.

Given:

The nitrogen gas molecule with a temperature of 330 Kelvins is released from Earth's surface to travel upward.

To find:

The maximum height of a nitrogen molecule when released from the Earth's surface before coming to rest.

Solution:

  • The maximum height attained by nitrogen gas molecule = h
  • The temperature of nitrogen gas particle = T = 330 K

The average kinetic energy of the gas particles is given by:

K.E=\frac{3}{2}K_bT\\\\K.E=\frac{3}{2}\times 1.38\times 10^{-23} J/K\times 330 K\\\\K.E=6.381\times 10^{-21} J

The nitrogen molecule at its maximum height will have zero kinetic energy as all the kinetic energy will get converted into potential energy

  • The potential energy at height h = P.E = 6.381\times 10^{-21} J
  • Molar mass of nitrogen gas =  28.0134 g/mol
  • Mass of nitrogen gas molecule = m

m= \frac{ 28.0134 g/mol}{6.022\times 10^{23} mol^{-1}}=4.652\times 10^{-23} g\\\\1g=0.001kg\\\\m=4.652\times 10^{-23}\times 0.001 kg\\\\=4.652\times 10^{-26} kg

  • The acceleration due to gravity = g = 9.8 m/s^2
  • The maximum height attained by nitrogen gas molecule = h
  • The potential energy is given by:

P.E=mgh

6.381\times 10^{-21} J=4.652\times 10^{-26} kg\times 9.8 m/s^2\times h\\\\h=\frac{6.381\times 10^{-21} J}{4.652\times 10^{-26} kg\times 9.8 m/s^2}\\\\h=13,996.6 m\\\\1 m = 0.001 km\\\\h=13,996.6 m=h=13,996.6\times 0.001 k m\\\\=13.9966 km \approx 14 km

The maximum height at which nitrogen molecule will go before coming to rest is 14 kilometers.

Learn more about the average kinetic energy of gas particles here:

brainly.com/question/16615446?referrer=searchResults

brainly.com/question/6329137?referrer=searchResults

6 0
2 years ago
24g of methane were burned in an excess of air. What mass of water would be produced in the reaction assuming complete combustio
expeople1 [14]

Answer:

54g of water

Explanation:

Based on the reaction, 1 mole of methane produce 2 moles of water.

To solve this question we must find the molar mass of methane in order to find the moles of methane added. With the moles of methane and the chemical equation we can find the moles of water produced and its mass:

<em>Molar mass CH₄:</em>

1C = 12g/mol*1

4H = 1g/mol*4

12g/mol + 4g/mol = 16g/mol

<em>Moles methane: </em>

24g CH₄ * (1mol / 16g) = 1.5 moles methane

<em>Moles water:</em>

1.5moles CH₄ * (2mol H₂O / 1mol CH₄) = 3.0moles H₂O

<em>Molar mass water:</em>

2H = 1g/mol*2

1O = 16g/mol*1

2g/mol + 16g/mol = 18g/mol

<em>Mass water:</em>

3.0moles H₂O * (18g / mol) =

<h3>54g of water</h3>
8 0
3 years ago
Neutral atoms of all the isotopes of the same element have​
Viktor [21]

Answer:

electron = proton

Explanation:

6 0
3 years ago
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