All categories

3 years ago

A covalent bond that is formed between two fluorine atoms has how many valence electrons shared between the two atoms?

Chemistry

1 answer:

valentinak56 [21]3 years ago

8 0

Two atoms die in the mode

You might be interested in

Which of the following is a heterogeneous mixture?

sesenic [268]

C because it is two seperate mixtures I think

5 0

4 years ago

Read 2 more answers

Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product o

stiv31 [10]

Answer:

Follows are the solution:

Explanation:

A + B = C

Its response decreases over time as well as consumption of a reactants.

r = -kAB

during response A convert into 2x while B convert into x to form 3x of C

let's y = C

y = 3x

Still not converted sum of reaction

for A: 100 - 2x

for B: 50 - x

Shift of x over time

$\frac{dx}{dt} = \frac{-k(100 - 2x)}{(50 - x)}$

Integration of x as regards t

$\frac{1}{[(100 - 2x)(50 - x)]} dx = -k dt\\\\\frac{1}{2[(50 - x)(50 - x)]} dx = -k dt\\\\\ integral\ \frac{1}{2[(50 - x)^2]} dx =\ integral [-k ] \ dt\\\\\frac{-1}{[100-2x]} = -kt + D \\\\$

D is the constant of integration

initial conditions: t = 0, x = 0

$\frac{-1}{[100-2x]} = -kt + D \\\\\frac{ -1}{[100]} = 0 + D\\\\D= \frac{-1}{100}\\\\$

hence we get:

$\frac{-1}{[100-2x]}= -kt -\frac{1}{100}\\\\or \\\\ \frac{1}{(100-2x)} = kt + \frac{1}{100}$

after t = 7 minutes , $C = 10 \ g = 3x$

$3x = 10\\\\x = \frac{10}{3}$

Insert the above value x into $\frac{1}{(100-2x)}$ equation $= kt + \frac{1}{100}$ to get k.

$\to \frac{1}{(100-2\times \frac{10}{3})} = k \times (7) + \frac{1}{100} \\\\ \to \frac{1}{(100- 2 \times 3.33)} = \frac{700k + 1}{100} \\\\ \to \frac{1}{(100-6.66)} = \frac{700k + 1}{100}\\\\ \to \frac{1}{93.34} = \frac{700k + 1}{100} \\\\$

$\to 100 = 93.34(700k + 1) \\\\ \to 100 = 65,338k + 700 \\\\ \to 65,338k = -600 \\\\ \to k = \frac{-600}{ 65,338} \\\\ \to k= - 0.0091$

therefore plugging in the equation the above value of k

$\to \frac{1}{(100-2x)} = kt +\frac{1}{100} \\\\\to \frac{1}{(100-2x)} = -0.0091t + \frac{1}{100}\\\\\to \frac{1}{(100-2x)} = \frac{1 -0.91t}{100}\\\\\to \frac{1}{2(50-x)} = \frac{1 -0.91t}{100}\\\\\to \frac{1}{(50-x)} = \frac{1 -0.91t}{50}\\\\\to 50= (1-0.91t)(50-x)\\\\\to 50 = 50-45.5t-x-0.91tx\\\\\to x+0.91xt= -45.5t\\\\\to x(1+0.91t)= -45.5t\\\\\to x=\frac{-45.5t}{1+0.91t}$

Let y = C

, calculate C:

y = 3x

$y =3 \times \frac{-45.5t}{1+0.91t}$

amount of C formed in 28 mins

$x = \frac{-45.5t}{1+0.91t} ,$ plug t = 28

$\to x = \frac{-1274}{1+25.48} \\\\\to x = \frac{-1274}{26.48} \\\\\to x= -48.26$

therefore amount of C formed in 28 minutes is = 3x = 144.78 grams

C: $y =3 \times \frac{-45.5t}{1+0.91t}$

y= 136.5 =137

7 0

3 years ago

Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons. Show the for

Alexus [3.1K]

Answer:

See explanation below

Explanation:

In this case, let's see both molecules per separate:

In the case of SeO₂ the central atom would be the Se. The Se has oxidation states of 2+, and 4+. In this molecule it's working with the 4+, while oxygen is working with the 2- state. Now, how do we know that Se is working with that state?, simply, let's do an equation for it. We know that this molecule has a formal charge of 0, so:

Se = x

O = -2

x + (-2)*2 = 0

x - 4 = 0

x = +4.

Therefore, Selenium is working with +4 state, the only way to bond this molecule is with a covalent bond, and in the case of the oxygen will be with double bond. See picture below.

In the case of CO₂ happens something similar. Carbon is working with +4 state, so in order to stabilize the charges, it has to be bonded with double bonds with both oxygens. The picture below shows.

5 0

4 years ago

Be sure to answer all parts. Solving the Rydberg equation for energy change gives ΔE = R[infinity]hc [ 1 n12 − 1 n22 ] where the

icang [17]

Answer:

Explanation:

Utilizing Rydber's equation:

ΔE = Z²Rh ( 1/n₁² - 1/n₂²) and substituting the values given ( using the Rydbers constant value in Joules ), we have

n=1 to n= infinity

ΔE = 3² x (1/1² - 0) x 2.18 x 10⁻^18 J = 2.0 x 10⁻¹⁷ J (1/infinity is zero)

n= 3 to n= infinity

ΔE = 3² x (1/3² - 0) x 2.18 x 10⁻^18 J = 2.28 x 10^-18 J

b. The wavelength of the emitted can be obtained again by using Rydberg's equation but this time use the constant value 1.097 x 10⁷ m⁻¹ given in the problem .

1/λ = Z²Rh (1/n₁² - 1/n₂²) 10 ⁻¹ = 3² x 1.097 x 10⁷ m⁻¹ x (1/1² - 1/3²) m⁻¹

1/λ =8.8 x 10⁷ m⁻¹ ⇒ λ =1.1 x 10^-8 m

λ = 1.1 x 10^-8 m x 1 x 10⁹ nm/m = 11 nm

7 0

3 years ago

Read 2 more answers

Which figure is a mixture ? #3

vovangra [49]

Answer:

Fig 3 is a mixture

Explanation:

Fig 1 has just one kind of atoms.

fig 2 shows 2 atoms of the same kind bonded together (compound)

fig 4 shows atoms of 2 different kinds bonded together (compound)

only in fig 3 we see uneven distribution of two kinds of atoms. By definition a mixture contains two or more kind of substances randomly arranged. So fig 3 shows arrangement of mixture

5 0

3 years ago