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3 years ago

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Which of the following has the largest radius? Answer:K

1 answer:

mixas84 [53]3 years ago

4 0

Answer:

the answer is k? lol thanks

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Why B-Cl bond is polar?

aniked [119]

Answer:

The difference in the electronegativities of chlorine and boron is 3.0 - 2.0 = 1.0 ; the difference in between chlorine and carbon is 3.0 = 2.5 = 0.5 . Consequently, the B-Cl bond is more polar ; the chlorine atom asrries the partial negative charge because it has higher electronegativity .

Explanation:

hope it helps!

8 0

3 years ago

Read 2 more answers

A certain reaction is thermodynamically favored at temperatures below 400. K, but it is not favored at temperatures above 400. K

Ilya [14]

Answer:

-0.050 kJ/mol.K

Explanation:

A certain reaction is thermodynamically favored at temperatures below 400. K, that is, ΔG° < 0 below 400. K

The reaction is not favored at temperatures above 400. K, that is. ΔG° > 0 above 400. K

All in all, ΔG° = 0 at 400. K.

We can find ΔS° using the following expression.

ΔG° = ΔH° - T.ΔS°

0 = -20 kJ/mol - 400. K .ΔS°

ΔS° = -0.050 kJ/mol.K

8 0

3 years ago

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Which household items are usually made of metal elements?

Rus_ich [418]

a=electronic equipment and jewelry

5 0

3 years ago

What experiments did Lavoisier do?

VikaD [51]

He did experiments with combustion and gas's

8 0

3 years ago

Practice entering numbers that include a power of 10 by entering the diameter of a hydrogen atom in its ground state, dH=1.06×10

kkurt [141]

Answer:

The diameter of the hydrogen $\mathbf{d =1.0605 \times 10^{-10}\ m}$

Explanation:

From the given information:

Using the concept of Bohr's Model, the equation for the angular momentum can be expressed as:

$L = \dfrac{nh}{2 \pi}$

Where the generic expression for angular momentum is:

L = mvr.

replacing the value of L into the previous equation, we have:

$mvr= \dfrac{nh}{2 \pi}$

$v= \dfrac{nh}{2 \pi mr}$ ----- (1)

The electron in the hydrogen atom posses an electrostatic force which gives a centripetal force.

$\dfrac{ke^2}{r^2} = \dfrac{mv^2}{r}$ ----- (2)

replacing the value of v in equation (1) into (2), and taking r as the subject of the formula, we have:

$\dfrac{ke^2}{r} = m (\dfrac{nh}{2 \pi mr})^2$

$ke^2=\dfrac{n^2h^2}{4 \pi^2 mr}$

$r =\dfrac{n^2h^2}{4 \pi^2 mke^2}$

For ground-state n = 1

$h = (6.625 \times 10^{-34} \ J.s)^2$

$m =( 9.1 \times 10^{-31} \ kg)(9 \times 10^9 \ N .m^2/C^2)$

$Ke = (1.6 \times 10^{-19} \ C)^2$

$r =\dfrac{(1)^2(6.625 \times 10^{-34})^2}{4 \pi^2 (9.1 \times 10^{-31} )(9 \times 10^9 ) (1.6 \times 10^{-19})^2}$

$r =\dfrac{4.3890625 \times 10^{-67}}{8.27720295 \times 10^{-57}}$

$\mathbf{r = 5.3025 \times 10^{-11} \ m}$

Therefore, the diameter of hydrogen d = 2r

$\mathbf{d = ( 2 \times 5.3025 \times 10^{-11} \ m})}$

$\mathbf{d =1.0605 \times 10^{-10}\ m}}$

4 0

3 years ago