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juin
3 months ago
8

According to the VSEPR model, the arrangement of electron pairs around NH3 and CH4 is A. the same, because in each case there ar

e the same number of electron pairs around the central atom B. different or the same, depending on the conditions leading to maximum repulsion C. different, because in each case there are a different number of atoms around the central atom D. the same, because both nitrogen and carbon are both in the second period E. different, because in each case there are a different number of electron pairs around the central atom
Chemistry
1 answer:
lorasvet [2.7K]3 months ago
3 0

Response:

Option "B" is the correct choice.

Explanation:

Based on VSEPR theory, repulsive forces occur between bond pair - bond pair or bond pair - lone pair of electrons. In both NH_{3} and CH_{4}, the electron pair count is the same, yet methane consists entirely of four bond pairs while ammonia contains three bond pairs and one lone pair. Therefore, repulsive forces are present between the lone and bond pairs of electrons, which causes the arrangement of electron pairs around the two molecules to vary depending on the circumstances to minimize repulsion.

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If angle ABE = 2n + 7 and angle EBF=4n-13,<br>find angle ABE.​
VMariaS [2998]

Answer:

Angle ABE measures 27°.

Explanation:

Refer to the attached diagram related to this question.

The given values are ∠ABE=2n+7 and ∠EBF=4n-13.

Clearly seen in the diagram, ∠ABE and ∠EBF are equal in measure.

m\angle ABE=m\angle EBF

2n+7=4n-13

Move variable components to one side of the equation.

7+13=4n-2n

20=2n

Split both sides by 2.

10=n

The solution for n arrives at 10.

The next step is to calculate ∠ABE.

\angle ABE=2(10)+7=20+7=27

Consequently, the measurement of angle ABE is 27°.

4 0
3 months ago
In a coffee-cup calorimeter experiment, 10.00 g of a soluble ionic compound was added to the calorimeter containing 75.0 g h2o i
VMariaS [2998]
Given: Mass of the ionic compound = 10.00 g Mass of water = 75.0 g Initial temperature of water T1= 23.2 C Final temperature of water T2 = 31.8 C Specific heat of water c = 4.18 J/gC To determine: Enthalpy of dissolution of the ionic compound Heat gained by water equation: Q = mcΔT m = mass of water c = specific heat ΔT = change in temperature (T2-T1) Q = 75.0 g * 4.18 J/gC * (31.8-23.2)C = 2696 J Thus, the heat gained by water equals heat lost by the ionic compound (enthalpy of dissolution) Therefore, q(ionic) = 2696 J ΔH = q(ionic)/mass of ionic compound = 2696 J/10.00 g = 2.7 *10² J/g Answer: A) enthalpy change = 2.7*10² J/g
7 0
2 months ago
Read 2 more answers
Does the result of the calculation in question 3 justify your original assumption that all of the SCN^- is in the form of FeNCS^
lorasvet [2795]
Fe 3+ + SCN- --> FeSCN 2+ 

<span>.......Fe 3+.......SCN-.........FeSCN 2+ </span>
<span>I.......0.04..........0.001.............. </span>
<span>C........-x...............-x............. </span>
<span>E.....0.04-x.....0.001-x...........x </span>

<span>Keq = 203.4 = x / (0.04-x)(0.001-x) </span>
<span>203.4 = x / (x^2 - 0.041x + 4x10^-5) </span>
<span>203.4x^2 - 8.34x + 0.00094 = x </span>
<span>203.4x^2 - 9.34x + 0.00094 = 0 </span>
<span>x = -0.0001M or 0.0458M </span>
<span>therefore, according to the calculated Keq, all of the SCN- and Fe 3+ would be fully converted into FeSCN 2+</span>
5 0
3 months ago
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