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2 years ago

Which of the following describes the electron sharing between hydrogen and fluorine?

3 answers:

4 0

Hello
We know both the elements involved in bonding are non-metals and the primary type of bonds involved in non-metals are covalent bonds. Covalent bonds are formed when two atoms share one or more electrons; thus we know that whatever the number of electrons shared, it has to be equal for both. We can eliminate choices A and B.
Next, we understand that it is easier for one atom to be shared among the two, rather than the 7. First, because Hydrogen needs only 1 electron to be stable and would require energy to compensate the remaining 6. Second, electrons are attracted towards the nucleus so it is inefficient to try and share 7 electrons when 1 provides the same amount of stability to Fluorine.
Therefore, the answer is C.

geniusboy [140]2 years ago

3 0

C. if you dont wantt ot read the whole paragraph it is C

Guest1 year ago

0 0

Which statement describes electron sharing between hydrogen and fluorine

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A particular reaction has an enthalpy and entropy of reaction of ∆H = +33 kJ/mol and ∆S = +0.15 kJ/mol⋅K. At the three indicated

Thepotemich [5.8K]

Answer:

Explanation: check my paage

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1 year ago

A 0.380 kg sample of aluminum (with a specific heat of 910.0 J/(kg x K)) is heated to 378 K and then placed in 2.40 kg of water

lbvjy [14]

Answer:

The equilibrium temperature of the system is 276.494 Kelvin.

Explanation:

Let consider the system formed by the sample of aluminium and water as a control mass, in which the sample is cooled and water is heated until thermal equilibrium is reached. The energy process is represented by First Law of Thermodynamics:

$Q_{water} -Q_{sample} = 0$

$Q_{water} = Q_{sample}$

Where:

$Q_{water}$ - Heat received by water, measured in joules.

$Q_{sample}$ - Heat released by the sample of aluminium, measured in joules.

Given that no mass is evaporated, the previous expression is expanded to:

$m_{w}\cdot c_{p,w}\cdot (T-T_{w}) = m_{s}\cdot c_{p,s}\cdot (T_{s}-T)$

Where:

$m_{s}$ , $m_{w}$ - Mass of water and the sample of aluminium, measured in kilograms.

$c_{p,s}$ , $c_{p,w}$ - Specific heats of the sample of aluminium and water, measured in joules per kilogram-Kelvin.

$T_{s}$ , $T_{w}$ - Initial temperatures of the sample of aluminium and water, measured in Kelvin.

$T$ - Temperature which system reaches thermal equilibrium, measured in Kelvin.

The final temperature is now cleared:

$(m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s})\cdot T = m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}$

$T = \frac{m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}}{m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s}}$

Given that $m_{s} = 0.380\,kg$ , $m_{w} = 2.40\,kg$ , $c_{p,s} = 910\,\frac{J}{kg\cdot K}$ , $c_{p,w} = 4186\,\frac{J}{kg\cdot K}$ , $T_{s} = 378\,K$ and $T_{w} = 273\,K$ , the final temperature of the system is:

$T = \frac{(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)\cdot (378\,K)+(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)\cdot (273\,K)}{(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)+(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)}$

$T = 276.494\,K$

The equilibrium temperature of the system is 276.494 Kelvin.

7 0

2 years ago

A solution at 25 ∘C has pOH = 10.53. Which of the following statements is or are true? The solution is acidic. The pH of the sol

Lady bird [3.3K]

Answer:

The true statements are:

The solution is acidic

The pH of the solution is 14.00 - 10.53.

$10^{-10.53}=[OH^-]$

Explanation:

The pH of the solution is defined as negative logarithm of hydrogen ion concentration present in the solution .

$pH=-\log[H^+]$

The pH value more 7 means that hydrogen ion concentration is less ,alkaline will be the solution.
The pH value less 7 means that hydrogen ion concentration is more ,acidic will be the solution.
The pH value equal to 7 indicates that the solution is neutral.

The pOH of the solution is defined as negative logarithm of hydroxide ion concentration present in the solution .

$pOH=-\log[OH^-]$

The pOH of the solution = 10.53

$10.53=-\log[OH^-]$

$10^{-10.53}=[OH^-]$

The pH of the solution = ?

$pH+pOH=14$

$pH=14-pOH=14-10.53=3.47$

Here, the pH of the solution is less than 7 which means that solution acidic.

8 0

2 years ago

A sample of gas initially occupies 3.35 L at a pressure of 0.950 atm at 13.0oC. What will the volume (in L) be if the temperatur

Marat540 [252]

Answer: V= 3.13 L

Explanation: solution attached:

Use combine gas law equation:

P1 V1 / T1 = P2 V2/ T2

Derive to find V2

V2 = P1 V1 T2 / T1 P2

Convert temperatures in K

T1= 13.0°C + 273 = 286 K

T2= 22.5°C + 273 = 295.5 K

Substitute the values.

4 0

2 years ago

When 100 ml of 1.0 M Na3PO4 is mixed with 100 ml of 1.0 M AgNO3,a yellow precipitate forms and Ag+ becomes negligibly small. Whi

jok3333 [9.3K]

Answer:

Option A

Explanation:

Number of millimoles of Na3PO4 = 1 × 100 = 100

Number of millimoles of AgNO3 = 1 × 100 = 100

When 1 mole of Na3PO4 is dissociated we get 3 moles of sodium ions and 1 mole of phosphate ion

When 1 mole of AgNO3 is dissociated, we get 1 mole of Ag+ and 1 mole of NO3-

As Ag+ concentration is negligible, the dissociated Ag+ ion must have form the precipitate with phosphate ion and as number of moles of Ag+ and phosphate ion are same, therefore the concentration of phosphate ion must be negligible

Here as 100 millimoles of Na3PO4 is there, we get 300 millimoles of Na+ and 100 millimoles of PO43-

And as 100 millimoles of AgNO3 is there, we get 100 millimoles of Ag+ and 100 millimoles of NO3-

∴ Increasing order of concentration will be PO43- < NO3- < Na+

4 0

2 years ago

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