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3 years ago

A potassium ion (K+) would most likely bond with?

Chemistry

1 answer:

ivanzaharov [21]3 years ago

6 0

It's most likely to bond with chlorine.

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A horse runs 1 km in 15 s. What is the average speed

Reptile [31]

The average speed of the horse for the entire motion is 66.67 m/s.

The given parameters;

<em>distance traveled by the horse, d = 1 km = 1,000 m</em>
<em>time of motion of the horse, t = 15 s</em>

The average speed of the horse is calculated by dividing the total distance traveled by the horse by the time of motion.

The average speed of the horse is calculated as follows;

$average\ speed = \frac{distance}{time} \\\\average\ speed = \frac{1000 \ m}{15 \ s} = 66.67 \ m/s$

Thus, the average speed of the horse for the entire motion is 66.67 m/s.

Learn more here: brainly.com/question/17289046

6 0

3 years ago

Which type of molecule contains two atoms bonded together

solmaris [256]

simple molecules contain two atom bonded together

5 0

3 years ago

Read 2 more answers

I’ll give brainliest!!!

vodka [1.7K]

Is this prier to a lab you've done?

4 0

4 years ago

Which property is true for metals?

Aleks04 [339]

They're malleable and have high melting and boiling points

8 0

4 years ago

Rank these transition metal ions in order of decreasing number of unpaired electrons.

lesya692 [45]

Answer: The given transition metal ions in order of decreasing number of unpaired electrons are as follows.

$Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}$

Explanation:

In atomic orbitals, the distribution of electrons of an atom is called electronic configuration.

The electronic configuration in terms of noble gases for the given elements are as follows.

Atomic number of Fe is 26.

$Fe^{3+} - [Ar] 3d^{5}$

So, there is only 1 unpaired electron present in $Fe^{3+}$ .

Atomic number of Mn is 25.

$Mn^{4+} - [Ar]3d^{3}$

So, there are only 3 unpaired electrons present in $Mn^{4+}$ .

Atomic number of V is 23.

$V^{3+} - [Ar] 3d^{2}$

So, there are only 2 unpaired electrons present in $V^{3+}$ .

Atomic number of Ni is 28.

$Ni^{2+} - [Ar] 3d^{8}$

So, there will be 2 unpaired electrons present in $Ni^{2+}$ .

Atomic number of Cu is 29.

$Cu^{+} - [Ar] 3d^{10}$

So, there is no unpaired electron present in $Cu^{+}$ .

Therefore, given transition metal ions in order of decreasing number of unpaired electrons are as follows.

$Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}$

Thus, we can conclude that given transition metal ions in order of decreasing number of unpaired electrons are as follows.

$Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}$

7 0

3 years ago