Using VSEPR model, how is the electron arrangement about the central atom (electron-pair geometry) for XeF4 ?

Answers

Answer 1

Answer:

See explanation

Explanation:

XeF4 has a square planar structure. There are six electron pairs on the valence shell of the central atom. This gives rise to an octahedral geometry based on valence shell electron pair repulsion theory.

There are two lone pairs of electrons present on the central atom, hence the four bond pairs are found to occupy the corners of a square thereby giving a square planar compound. The two lone pairs are found above and blow the plane of the square.


Related Questions

Round off the following number to four significant figures.
19.9743

Answers

Answer:

4 sig fis so look at the next number which is "4"

answer is 19.97

all numbers not zeros are significant

Based on the results of your titration, what volume of the sample would an adult need to consume to reach the recommended daily amount? 60 mg is the recommened daily amount of vitamin C.

Ascorbic acid Volume of DCPIP in Total Volume of DCPIP

concentration (g/L) final syringe (mL) used (mL)

0.00 2.18 mL 0.18 mL

0.50 3.1 mL 0.6 mL

1.00 4.6 mL 2.2 mL

1.50 5.56 mL 3.39 mL

2.00 6.37 mL 4.37 mL

Unknown Trial 1 3.21 mL 1.21 mL

Unknown Trial 2 3.11 mL 1.11 mL

Unknown Trial 3 3.15 mL 1.06 mL

Answers

Answer:  85.7 mL

Explanation:

Given the information from the question as plotted in the graph i will be uploading along side this answer,

Average of total volume of DCPIP used is

= (1.21 + 1.11 + 1.06)mL / 3

= 1.12 mL

and corresponding ( ascorbic acid ) is 0.70 g/L

Two parameter given as volume of DCPIP in final syringe and total volume of DCPIP are quite ambiguous

700mg ⇒ 1 L

THEREFORE volume that contains 60mg =  (1000/700) × 60 = 85.7 mL

Periodic Table High School Reference Sheet
Which of the following equations best describes the energy at point
C, assuming energy is conserved?
A PEA = KEC
B
PEA = KEC - PEC
C. PEB = KEO - PEO
D
PES = PEC

Answers

I believe it should be B

Given that blood exerts the same osmotic pressure as a 0.15 M NaCl solution, which solution could be the isotonic solution

Answers

Answer:

0.15 (M) of NaCl            

Explanation:

An isotonic solution is defined as the solution having the same osmolarity, or the same solute concentration, as the another solution. If the two solutions are to be separated by any semipermeable membrane, then water will start flowing in equal parts from each solution and also into the other.  

In the context, since it is given that the red blood cells exerts the same osmotic pressure as a 0.15 (M) of NaCl solution.

Thus, 0.15 (M) of NaCl is a isotonic solution.  

How much is the chemotherapy in FeS2?explain

Answers

Answer:

Valence of iron in FeS2

Let the valency of iron be x

x+2(oxidation number of Sulphur)=0 as their is no charge on FeS2

As sulphur belongs to Group VI (chalcogens) so the valency/oxidation number of Sulphur is -2

x+2(-2)=0

x-4=0

x=4

So the valency/oxidation number of iron in FeS2 is +4

Explanation:

Hope it will help you monkey ^_^ it's crab

Answer:

Hope you are better now your parents must be worried about you get well soon  drink plenty of milk it will make your bones strong.

NOW TIME FOR WISHES ^_^

Approximately how many moles of Al3 are reduced when 0.1 faraday of charge passes through a cell during the production of Al

Answers

Answer:

HELLO YOUR QUESTION IS INCOMPLETE ATTACHED BELOW IS THE COMPLETE QUESTION

answer : 0.033 mol  ( A )

Explanation:

To determine the number of moles of AI3 that will be reduced when 0.1 Faraday passes through a cell during the production of AI

AI3+  + 3 e -------- AI  this means that it will take 3 mol to reduce AI3 to AI

1 mol AI = 3 F  electrons

therefore : 1 F electrons  = 1/3 mol of AI

therefore : 0.1 F ( Faraday ) = 0.1 * 1/3  = 0.033 mol of AI is needed

determine the ratio of the diatomic element with equal percentage abundance​

Answers

Answer:

I ratio is

47 : 43 ratio

carbon dioxide at 25 degrees celsius and 101.3 kpa has a density of 1.799 kg/m^3 determine gas constant

Answers

Answer:

Explanation:

Temperature of gas in absolute scale T = 25 + 273 = 298 .

pressure of gas P = 1.013 x 10⁵ N / m²

density D = 1.799 kg / m³

= 1799 g / m³

From gas formula

PV / T = n R Where P is pressure , V is volume and T is absolute temperature , n is no of moles

P / T = n R / V

P / T = m R /M V where m is mass of gas and M is molecular weight .

m / V = D ( density )

P / T = DR/ M  

PM / DT = R

Putting the values

1.013 x 10⁵ x 44  /  (1799 x 298)

R = 8.314.09 J / K mole

A 2.0 mL sample of carbon tetrachloride (CCI4 ) masses at 3.16 g. What is the density of this substance in g/mL?
A. 1.5 g/mL
B. 1.58 g/mL
C. 1.6 g/ml
D. 0.63 g/ml

Answers

Answer:

The answer is option B

Explanation:

The density of a substance can be found by using the formula

[tex]Density = \frac{mass}{volume} [/tex]

From the question

mass of CCl4 = 3.16 g

volume of CCl4 = 2.0 mL

Substitute the values into the above formula and solve for the Density

That's

[tex]Density = \frac{3.16}{2} [/tex]

We have the final answer as

Density = 1.58 g/mL

Hope this helps you

30.0 mL of pure water at 282 K is mixed with 50.0 mL of pure water at 328 K. What is the final temperature of the mixture

Answers

Answer:

Explanation:

As the 50mL of water are in a big temperature will give heat to the other sample of water.

With the equation:

Q = m*C*T

Where Q is heat, m  is mass of the sample, C is specific heat of the sample (4.184J/molK for pure water) and T is temperature in K

We can determine the energy of each sample of water:

Sample 1:

Q = 30.0g*4.184J/molK*282K

Mass is 30.0g because density of water is 1g/mL

Q = 35396.6J

Sample 2:

Q = 50.0g*4.184J/molK*328K

Q = 68617.6J

Total energy is:

Q = 104014.2J

As total mass is 30+50 = 80g, final temperature is:

104014.2J = 80.0g*4.184J*T

310.7K is final temperature of the mixture

How many types of ions must be present in solution, at minimum, for the phenomenon of selective precipitation to be observed? Your answer should be a whole number without any decimal places.

Answers

Answer:

Two (2) types of ions

Explanation:

Metal ions (cations) and negatively charged ions (anions) of either chloride, carbonate, sulphides or hydroxides are present during the selective precipitation.

Majorly, the process determines the metal ions present by initiating precipitation as certain conditions.

I hope this was helpful.

The average adult heart pumps about 84./mLs of blood at 72 beats per minute. Suppose you need to calculate how long it would take the average heart to circulate 3600.mL of blood. Set the math up. But don't do any of it. Just leave your answer as a math expression. Also, be sure your answer includes all the correct unit symbols.

Answers

Answer:

42.857 seconds

Explanation:

Given that:

The average adult heart pumps about 84./mLs  of blood at 72 beats per minute.

The total volume = 3600 mL

The objective is to calculate how long it would take the average heart circulate 3600 mL of blood.

Since 84 mL of blood takes 1 seconds to pump,

assuming q is the time required for heart to pump 3600 mL of blood

[tex]Time \ ( q )= \dfrac{3600 mL \times 1\ seconds}{84 \ mL}[/tex]

TIme (q) = 42.857 seconds

Calculate the mass of the air contained in a room that measures 2.50 m × 5.50 m × 3.00 m (density of air = 1.29 g/dm3 at 25°C).

Answers

Answer:

Explanation:

Volume of air = 2.5 x 5.5 x 3.00

= 41.25 m³.

Density of air = 1.29 g / dm³

a dm³ = ( 0.1 )³m³

= .001 m³

density in m³ = 1.29 / .001 gm⁻³

= 1290 g / m³

mass of air in the room

= 41.25 x 1290 g

= 53212.5 g

= 53.2125 kg .

Explanation:

please mark me as brainlest

A 0.0678 M solution of a weak base has a pH of 9.44. What is the identity of the weak base? Weak Base Kb Ethylamine (CH3CH2NH2) 4.7 × 10–4 Hydrazine (N2H4) 1.7 × 10–6 Hydroxylamine (NH2OH) 1.1 × 10–8 Pyridine (C5H5N) 1.4 × 10–9 Aniline (C6H5NH2) 4.2 × 10–10

Answers

Answer:

The weak base is hydroxylamine

Explanation:

The general equilibrium of a weak base (A), is:

A(aq) + H₂O(l) ⇄ HA⁺ + OH⁻

Where Kb is:

Kb = [HA⁺] [OH⁻] / [A]

As both HA⁺ and OH⁻ comes from the equilibrium of A; HA⁺ = OH⁻. So, you can take:

Kb = [OH⁻]² / [A]

OH⁻ is determined from pH, thus:

pOH = -log [OH⁻]

pOH = 14 - pH

pOH = 14 - 9.44

pOH = 4.56

[OH⁻] = 10^{-4.56}

[OH⁻] = 2.75x10⁻⁵

As the concentration of the weak base is 0.0678M:

Kb = [OH⁻]² / [A]

Kb = [2.75x10⁻⁵]² / [0.0678M]

Kb = 1.1x10⁻⁸ = Kb of Hydroxylamine.

The weak base is hydroxylamine

can anyone help me with nomenclature?​

Answers

Explanation:

a chemical nomenclature is a set of rules to generate systematic names for chemical compounds.

A sample initially contains 4.4 moles of a radioactive isotope. How much of the sample remains after four half-lives?

Answers

Answer:

0.275

Explanation:

Given that:

Initial number of miles of radioactive isotope = 4.4 moles

How much if the sample remains after 4 half life's

Half-life refers to the time taken by a sample to reduce to half of it's initial value.

After the first half life :

Initial / 2 = 4.4 / 2 = 2.2 moles

Second half life :

2.2 / 2 = 1.1 moles

Third half life :

1.1 / 2 = 0.55 Moles

Fourth half life :

0.55 / 2 = 0.275 moles

Calculate the density in g/mL of 2.0 L of gasoline that weighs 1.34 kg

Answers

Answer:

Density = 0.67 g/mL

Explanation:

The density of a substance can be found by using the formula

[tex]Density = \frac{mass}{volume} [/tex]

To calculate the Density in g/mL we must first convert the mass from kg to g and the volume from L to mL

For the mass

That's

1.34 kg

1kg = 1000 g

1.34 kg = 1000 × 1.34 = 1340 g

For the volume

That's

2.0 L

1 L = 1000 mL

2.0 L = 2 × 1000 = 2000 mL

So we have

mass = 1340 g

volume = 2000 mL

Substitute the values into the above formula and solve for the Density

That's

[tex]Density = \frac{1340}{2000} \\ = \frac{67}{100} [/tex]

We have the final answer as

Density = 0.67 g/mL

Hope this helps you

Tracy had seen her teacher perform the demonstration many times. The materials for the demonstration were lying on the teacher’s table. Tracy tied her long hair back and used the materials to perform the demonstration on her own. a) Write an action that Tracy did correctly. b) Write an action that Tracy did not follow.

Answers

Answer:

Tracy was able to remember what her teacher told her to do and she decided to try it,which was the right thing.

But since it's a demonstration then Tracy is supposed to wear all her protective equipments and with a teacher's permission.I hope this helps.Thanks

(5.391 × 10−7) + (2.5531 × 10−6)

Answers

Answer:

66.441

I hope this helps!

Answer:

66.441

Explanation:

(5.391 × 10-7) + (2.5531 × 10-6)

(46.91) + (19.531)

66.441

You have been injured in the laboratory (cut,burn,etc) first you should

Answers

Let the teacher know

Answer:tell the instructor of the lab

Explanation:

Blake is eliminating contributing factors to ensure accuracy in his results. Which step of the scientific method is he performing?
O Research the hypothesis
O Research the problem
O Test the hypothesis
O Analyze the results
Conclusion

Answers

Answer:

C

Explanation:

The answer is letter c

"Splotchy" most likely refers to a mineral's:
O color
luster
streak
O hardness

Answers

Answer:Luster

Explanation: i'm tired so it is luster because splotchy wouldn't be referring to the others so it would be referring to Luster the reflection of light of the object. Hope it helps

PLZ BRAINLILEST I ONLY NEED ONE MORE

Answer:

the answer is color

Explanation:

i got it right on my assignment lol

What is the change in enthalpy when 9.00 mol of sulfur trioxide decomposes to sulfur dioxide and oxygen gas

Answers

Answer:

-891kj

Explanation:

We have the chemical equation as:

2SO2(g) + O2(g) ----2SO3; ∆H° = 198kj

If sulphur trioxide decomposes we have opposite reaction. Enthalpy of decomposition will not be positive for forward reaction.

2 moles of SO3 ----∆H° = -198Kj

1 mole = -198/2

9 moles = -198/2 x 9

= -891kj

Therefore change in enthalpy when 9 moles is decomposes is -891kj.

What types of knowledge and background did Underwood and Prescott bring to the problem?

Answers

Answer:

The answer is on the screenshot

Explanation:

A student graphed distance versus time for an object that moves 14 every 2 s. What is the speed of the object?

Answers

Answer:

The speed is 7 every second.

Equation:

14/2 = 7

A student graphed distance versus time for an object that moves 14 every 2 s. The speed of the object is 7 m/sec.

What is speed?

The speed is the distance traveled by an object with a specific unit of time. Speed is calculated by dividing distance by time. The SI unit of speed is a meter per second.

S = d / t

Distance is the space or extent between two things. It is measured in meters and kilometers.

Time is a variable that equals the distance divided by time.

The slope of a graph is the rate at which the y and x axis change together, the "rise over the run." Slope can be simply calculated using this equation:

Given, that distance = 14 meters

The time is 2 sec.

By the formula

d = 14 / 2 = 7 m /sec.

Thus, the speed of the object is 7 m/sec.

To learn more about speed, refer to the link:

https://brainly.com/question/7359669

#SPJ2

An element has three stable isotopes of masses 24, 25, and 26. They exist in abundances of 79%, 10%, and 11%, respectively. What is the average atomic mass, and what is the element? *

Answers

Answer:

12 this is your answer

hope you like it

A pool is 60.0 m long and 35.0 m wide. How many mL of water are needed to fill the pool to an average depth of 6.35 ft? Enter your answer in scientific notation.

Answers

Answer:

4.07 × 10⁹ mL

Explanation:

Step 1: Given data

Length of the pool (L): 60.0 m

Width of the pool (W): 35.0 m

Depth of water in the pool (D): 6.35 ft

Step 2: Convert "D" to m

We will use the relationship 1 m = 3.28 ft

[tex]6.35ft \times \frac{1m}{3.28ft} = 1.94m[/tex]

Step 3: Calculate the volume of water (V)

We will use the following expression.

[tex]V = L \times W \times D = 60.0m \times 35.0m \times 1.94 m = 4.07 \times 10^{3} m^{3}[/tex]

Step 4: Convert "V" to mL

We will use the relationship 1 m³ = 10⁶ mL.

[tex]4.07 \times 10^{3} m^{3} \times \frac{10^{6}mL }{1m^{3} } = 4.07 \times 10^{9} mL[/tex]

Phosphorus-32 is a commonly used radioactive nuclide in biochemical research, particularly in studies of nucleic acids. The half-life of phosphorus-32 is 14.3 days. What mass of phosphorus-32 is left from an original sample of 175 mg of Na₃³²PO₄ after 45.6 days? Assume that the atomic mass of ³²P is 32.0.

Answers

Answer:

The correct answer is 6.65 mg.

Explanation:

Based on the given information, the number of days given is 45.6 days. The original mass of sodium phosphate give is 175 mg. The half-life of phosphorus-32 is 14.3 days, therefore, the n or the number of half life will be,

n = 45.6/14.3 = 3.19

So, after 45.6 days, the mass of sodium phosphate sample left will be,

= Original mass × 1/2ⁿ

= 175 mg × 1/2 ^3.19

= 175 mg × 0.1096

= 34.3 mg of sodium phosphate left after 45.6 days

The molecular mass of Na₃³²PO₄ is 3 (23) + 32 + 4 (16) amu = 165 amu

Therefore, 165 grams of sodium phosphate comprise 32 grams of phosphorus.

So, 34.3 mg or 0.343 gram of sodium phosphate will contain,

= 0.343 × 32/165

= 6.65 mg of P³².

One of the differences between a voltaic cell and an electrolytic cell is that in an electrolytic cell __________.

Answers

Answer:

a nonspontanenous reaction is forced to occur

Hope this helps :)

how many grams of iron(ii) sulfate heptahydrate is needed to prepare 250.0 ml of a 3.25 molar aqueous solution

Answers

Answer:

The mass of FeSO₄.7H₂O  needed  [tex]\simeq[/tex] 225.89 grams

Explanation:

From the given question

The molecular weight of iron(ii) sulfate heptahydrate FeSO₄.7H₂O = 278.02 g/mol

Molarity = 3.25

Volume = 250.0 mL

The mass of FeSO₄.7H₂O  needed = [tex]\dfrac{moles \times molecular weight \times 250\ mL}{1000 \ mL}[/tex]

The mass of FeSO₄.7H₂O  needed  = [tex]\dfrac{3.25 \times 278.02 \times 250}{1000}[/tex]

The mass of FeSO₄.7H₂O  needed  = 225.89125 grams

The mass of FeSO₄.7H₂O  needed  [tex]\simeq[/tex] 225.89 grams

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