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arsen [322]
2 years ago
10

PLEASE HELP, I REALLY NEED HELP WITH THIS!!!​

Chemistry
1 answer:
Mkey [24]2 years ago
3 0

Answer:

Explanation:

There are 6 32S atoms and 2 36S atoms

Their respective contributions to the average aomic mass is (6*32)=192, and (2*36)=72.  Added together, this makes 264 total.  Divide by the total number of atoms, 8.  (264/8) = 33.00

The weighted atomic mass of these S atoms is 33.00 amu (atomic mass units)

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The element in period 3 with the most metallic character is A sodium
Gekata [30.6K]
A. Sodium is correct.

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8 0
3 years ago
Complete these structures by adding electron dots as needed.
LiRa [457]
This is hard to show but here is how you would determine these. NOTE each dot is an electron. 
<span>Question 1) </span>
<span>F-H </span>
<span>1) determine the valance electrons for each. F has 7 and H has 1 </span>
<span>2) one electron from both F and H form the bond "-" which means that you still have 6 electrons to place around F and none to place around H. Place the 6 in sets of 2 around the F </span>
<span>.. </span>
<span>F-H </span>
<span>¨ </span>
<span>Question 2) </span>

<span>2) H-O-H </span>
<span>H has 1 valence electron minus 1 used in the bond to O = 0 electrons to place </span>
<span>H has 1 valence electron minus 1 used in the bond to O = 0 electrons to place </span>
<span>O has 6 valence electrons minus 2 used in the bonds to the H's = 4 electrons to place </span>

<span>H-O-H: place two dots above and below the oxygen </span>

<span>Question 3) </span>
<span>3) O=N----H : NOTE: a double bond requires O and N to share two of their electrons each </span>
<span>O has 6 valence electrons minus 2 used in the bonds to N = 4 electrons to place </span>
<span>N has 5 valence electrons minus 3 used in the bonds to O and H = 2 electrons to place </span>
<span>H has 1 valence electron minus 1 used in the bond to N = 0 electrons to place </span>
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7 0
3 years ago
Consider the following reaction: CH3OH(g)⇌CO(g)+2H2(g) Part A Calculate ΔG for this reaction at 25 ∘C under the following condit
kati45 [8]

<u>Answer:</u> The \Delta G of the reaction at given temperature is -12.964 kJ/mol.

<u>Explanation:</u>

For the given chemical reaction:

CH_3OH(g)\rightleftharpoons CO(g)+2H_2(g)

The expression of K_p for the given reaction:

K_p=\frac{(p_{CO})\times (p_{H_2}^2)}{p_{CH_3OH}}

We are given:

p_{CO}=0.140atm\\p_{H_2}=0.180atm\\p_{CH_3OH}=0.850atm

Putting values in above equation, we get:

K_p=\frac{(0.140)\times (0.180)^2}{0.850}\\\\K_p=5.34\times 10^{-3}

To calculate the Gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_p

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = 0 J (at equilibrium)

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 5.34\times 10^{-3}

Putting values in above equation, we get:

\Delta G=0+(8.314J/K.mol\times 298K\times \ln(5.34\times 10^{-3}))\\\\\Delta G=-12963.96J/mol=-12.964kJ/mol

Hence, the \Delta G of the reaction at given temperature is -12.964 kJ/mol.

5 0
3 years ago
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Ede4ka [16]
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Read 2 more answers
PLZ HELP ME WITH MY WORK​
Delvig [45]

Answer:

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