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wel
4 years ago
13

TYPE OUT EXTREMELY NEATLY PLEASE AND SHOW ALL YOUR WORK. Use significant figures where appropriate

Chemistry
1 answer:
coldgirl [10]4 years ago
5 0

<u>Answer:</u> The number of moles of nitrogen gas is 9.9 moles.

<u>Explanation:</u>

To calculate the mass of bromine gas, we use the ideal gas equation, which is:

PV = nRT

where,

P = Pressure of nitrogen gas = 2.30 atm

V = Volume of nitrogen gas = 120.0 L

n = Number of moles of nitrogen gas = ? mol

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = Temperature of nitrogen gas = 340 K

Putting values in above equation, we get:

2.30atm\times 120.0L=n\times 0.0821\text{L atm }mol^{-1}K^{-1}\times 340K\\\\n=9.88mol\approx 9.9mol

Rule of significant figures in case of multiplication and division:

The least number of significant figures in any number of the problem will determine the number of significant figures in the solution.

Here, the least precise number of significant figures are 2. Thus, the number of moles of nitrogen gas is 9.9 moles.

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Ch3cooh (acetic acid) can form hydrogen bonds between its molecules. based on the lewis structure shown below, how many hydrogen
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Answer:
           Number of Hydrogen Bond Acceptor atoms  =  2

           Number of Hydrogen Bond Donor atoms  =  1

Explanation:
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                    In Acetic acid there are two oxygen atoms hence there are two most electronegative elements therefore, two Hydrogen Bond Acceptor atom and each oxygen atom can accept two hydrogen bonds.
                    Also, it contains only one Hydrogen atom attached to oxygen atom so it has one Hydrogen Bond Donor atom.

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If an acid or alkali is diluted by a factor of 10, what effect will this have on its pH?
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7 0
3 years ago
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Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Cr2 ] = 0.892 M and [Fe2 ] = 0.0150
Effectus [21]

<u>Given:</u>

Concentration of Cr2+ = 0.892 M

Concentration of Fe2+ = 0.0150 M

<u>To determine:</u>

The cell potential, Ecell

<u>Explanation:</u>

The half cell reactions for the given cell are:

Anode: Oxidation

Cr(s) ↔ Cr2+(aq) + 2e⁻                E⁰ = -0.91 V

Cathode: Reduction

Fe2+ (aq) + 2e⁻ ↔ Fe (s)              E⁰ = -0.44 V

------------------------------------------

Net reaction: Cr(s) + Fe2+(aq) ↔ Cr2+(aq) + Fe(s)

E°cell = E°cathode - E°anode = -0.44 - (-0.91) = 0.47 V

The cell potential can be deduced from the Nernst equation as follows:

Ecell = E°cell - (0.0591/n)log[Cr2+]/[Fe2+]

Here, n = number of electrons = 2

Ecell = 0.47 - 0.0591/2 * log[0.892]/[0.0150] = 0.418 V

Ans: The cell potential is 0.418 V

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3 years ago
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