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likoan [24]
2 years ago
15

At the melting point of a substance, temperature ______ as heat is being added and the substance is changing from a solid to a l

iquid.
A. increases

B. remains constant

C. decreases

D. ceases to exist
Chemistry
2 answers:
madam [21]2 years ago
4 0
The temperature Increases
timofeeve [1]2 years ago
4 0
At the melting point of a substance, temperature INCREASES as heat is being added and the substance is changing from a solid to a liquid
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Using the molecular orbital model to describe the bond- ing in F2????, F2, and F2????, predict the bond orders and the relative
masya89 [10]

Answer: F2 : bond order= 1.0

F2+: bond order = 1.5

F2- : bond order = 0.5

Explanation:

1. Starting with F2+

The configuration gives;

F2+ = 9F = 1S2.2S2.2P5

= 9F+ = 1S2.2S2.2P4 (this shows it gives out an electron)

Electronic configuration = σ 1S2σ*1S2σ2S2σ*2S2σ2Pz2 π2Px2 = π2Py2 π*2Px2 = π*2Py1

The number of Electrons = (9*2) – 1 = 18 -1 = 17

Bonding electrons (Nb) = 10

Antibonding electrons (Na) = 7

Bond order = (10-7)/2 = 3/2 = 1.5

Number of unpaired electrons = 1

2. Starting with F2

The configuration gives;

F2 = 9F = 1S2.2S2.2P5

9F = 1S2.2S2.2P5 (this shows no loss of an electron)

Electronic configuration = σ 1S2σ*1S2σ2S2σ*2S2σ2Pz2 π2Px2 = π2Py2 π*2Px2 = π*2Py2

The number of Electrons = (9*2) = 18 electrons

Bonding electrons (Nb) = 10

Antibonding electrons (Na) = 8

Bond order = (10-8)/2 = 2/2 = 1.0

Number of unpaired electrons = 0

3. Starting with F2-

The configuration gives;

F2- = 9F = 1S2.2S2.2P5

10F--= 1S2.2S2.2P6 (this shows an addition of an electron)

Electronic configuration = σ 1S2σ*1S2σ2S2σ*2S2σ2Pz2 π2Px2 = π2Py2 π*2Px2 = π*2Py2 σ*2Pz

The number of Electrons = (9*2) + 1 = 19 electrons

Bonding electrons (Nb) = 10

Antibonding electrons (Na) = 9

Bond order = (10-9)/2 = 1/2 = 0.5

Number of unpaired electrons = 1

To get the order of bond as well as length, we know that;

Bond order directly proportional to 1/ Bond length

Therefore the Ascending Bond length = F2+ ˂ F2 ˂ F2-

3 0
2 years ago
FIST ONE TO AWNSER GETS BRAINLYIST AND 50 POINTS IF CORRECT HURRY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
grin007 [14]

Answer:

B

Explanation:

Shows actual sacrifice in order to train and be the best.

7 0
3 years ago
Read 2 more answers
Is oxygen gas (O2) a compound or an element? Is hydrogen gas (H2) a compound or an element?
Alinara [238K]
They're both elements
4 0
3 years ago
Nitrogen reacts with hydrogen to produce ammonia gas as follows. mc023-1.jpg How many moles of nitrogen would react with excess
anzhelika [568]
The balanced reaction is:

N2 + 3H2 = 2NH3

We are given the amount of the product to be produced.This will be the starting point of our calculations. We use the ideal gas equation to find for the number of moles.

<span>
n = PV / RT = 1.00(.520 L) / (0.08206 atm L/mol K ) 273 K
n= 0.0232 mol NH3

</span>0.0232 mol NH3 (1 mol N2 / 2 mol NH3) = 0.0116 mol N2


<span>Therefore, the correct answer is A.</span>

7 0
3 years ago
Read 2 more answers
Suppose that 0.1000 mole each of H2and I2are placed in a 1.000-L flask, stoppered, and the mixture is heated to 425oC. At equili
Katen [24]

<u>Answer:</u> The value of equilibrium constant for the given reaction is 56.61

<u>Explanation:</u>

We are given:

Initial moles of iodine gas = 0.100 moles

Initial moles of hydrogen gas = 0.100 moles

Volume of container = 1.00 L

Molarity of the solution is calculated by the equation:

\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}

\text{Molarity of iodine gas}=\frac{0.1mol}{1L}=0.1M

\text{Molarity of hydrogen gas}=\frac{0.1mol}{1L}=0.1M

Equilibrium concentration of iodine gas = 0.0210 M

The chemical equation for the reaction of iodine gas and hydrogen gas follows:

                         H_2+I_2\rightleftharpoons 2HI

<u>Initial:</u>                0.1    0.1

<u>At eqllm:</u>          0.1-x   0.1-x   2x

Evaluating the value of 'x'

\Rightarrow (0.1-x)=0.0210\\\\\Rightarrow x=0.079M

The expression of K_c for above equation follows:

K_c=\frac{[HI]^2}{[H_2][I_2]}

[HI]_{eq}=2x=(2\times 0.079)=0.158M

[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M

[I_2]_{eq}=0.0210M

Putting values in above expression, we get:

K_c=\frac{(0.158)^2}{0.0210\times 0.0210}\\\\K_c=56.61

Hence, the value of equilibrium constant for the given reaction is 56.61

6 0
3 years ago
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