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madreJ [45]
3 years ago
13

How much work (in JJ) is required to expand the volume of a pump from 0.0 LL to 2.5 LL against an external pressure of 1.1 atmat

m
Chemistry
1 answer:
stira [4]3 years ago
8 0

Answer:

- 278.85 J  

Explanation:

Given that:

Pressure = 1.1 atm

The initial volume V₁ = 0.0 L

The final volume V₂ = 2.5 L

The work that takes place in a reaction at constant pressure can be expressed by using the equation:

W = P(V₂ - V₁ )

Since the volume of the gas is expanded from 0 to 2.5 L when 1.1 atm pressure is applied. Then, the work can be given by the expression:

W = - P(V₂ - V₁ )

W = -1.1 atm ( 2.5 - 0.0) L

W = -1.1 atm (2.5 L)

W = -2.75 atm L

Recall that:

1 atm L = 101.4 J

Therefore;

-2.75 atm L = ( -2.75 × 101.4 )J

= -278.85 J  

Thus, the work required at the chemical reaction when the pressure applied is 1.1 atm  = - 278.85 J  

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<span>Well it depends on percentage by what, but I'll just assume that it's percentage by mass. For this, we look at the atomic masses of the elements present in the compound. Cu has an atomic mass of 63.546 amu Fe has 55.845 amu and S has 36.065 amu Since there are 2 molecules of Sulfur for each one of Cu and Fe, we'll multiply the Sulfur atomic weight by 2 to obtain 72.13 amu So we have not established the mass of the compound in amus 63.546 + 55.845 + 72.13 = 191.521 That is the atomic mass of Chalcopyrite. and Iron's atomic mass is 55.845 So to get the percentage, or fraction of iron, we take 55.845 / 191.521 Which comes out to 29.15% by mass Mass of the sample is not needed for this calculation, but since the question mentions it I would go ahead and check if the question isn't also asking for the mass of Iron in the sample as well, in which case you just find the 29.15% of 67.7g</span>
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3 years ago
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pav-90 [236]
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3 years ago
Draw a bond-line structure for CH3CH2O(CH2)2CH(CH3)2.<br><br> Include Lone Pairs in your answer.
mash [69]

Answer:

See explanation and image attached

Explanation:

A bond line structure refers to any structure of a covalent molecule wherein the covalent bonds present in the molecule are represented with a single line for each level of bond order.

The bond-line structure of CH3CH2O(CH2)2CH(CH3)2 has been shown in the image attached. We know that oxygen has a lone pair of electrons and this has been clearly shown also in the image attached.

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3 years ago
How many atoms are in 8.28 moles of aluminum?
Lilit [14]

Answer:

49.86 × 10²³  atoms of Al

Explanation:

Given data:

Number of moles of Al = 8.28 mol

Number of atoms = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

For 8.28 moles of Al:

1 mole = 6.022 × 10²³  atoms of Al

8.28 mol×6.022 × 10²³  atoms / 1mol

49.86 × 10²³  atoms of Al

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