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3 years ago

Whos good at the periodic table cus im not help

Chemistry

2 answers:

just olya [345]3 years ago

6 0

Yep it’s 35.45 aka Chlorine. So C

Nesterboy [21]3 years ago

5 0

The answer should be C. 35

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What is neutralisation reaction? why is it named so? give one example.

docker41 [41]

Answer:

In chemistry, neutralization or neutralisation is a chemical reaction in which acid and a base react quantitatively with each other. In a reaction in water, neutralization results in there being no excess of hydrogen or hydroxide ions present in the solution.Neutralization reactions are the reaction between acid and base. The products formed are water and salt. It is called so because the acid and base neutralize each other to form water and salt.Hint: The neutralization reaction is the one in which an acid reacts with an equimolar amount of base to give salt and water. The example could be a reaction between any strong acid and a base. The sodium chloride formed is a result of neutralization reaction.

Trust me mark me as brainliest trust me

4 0

2 years ago

The oxidation number of Na in NaCl a. 0 b. -1 c. +1 d. -2 e. +2

Dahasolnce [82]

NaCl:

Na = + 1

Cl = - 1

hope this helps!

4 0

3 years ago

What kind of property is melting point? an atomic property , , , an atomic property , , a physical property , a physical propert

jok3333 [9.3K]

Answer:

physical property has a melting point

8 0

2 years ago

Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find

FromTheMoon [43]

Answer: The concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

Explanation:

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

$H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)$

0.025 0.025 0.025

Equation for the second dissociation of sulfuric acid:

$HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)$

Initial: 0.025 0.025

At eqllm: 0.025-x 0.025+x x

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}$

We know that:

$Ka_2\text{ for }H_2SO_4=0.01$

Putting values in above equation, we get:

$0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

4 0

3 years ago

Pleaseee helpppppppp

ahrayia [7]

Answer:

Potassium

1s2 2s2 2p6 3s2 3p6 4s1

Explanation:

The atom having only one electron its outermost shell must belong to an element in group one of the periodic table.

Having noted that, we proceed to find out what element in group one that has the atom just described in the question.

That atom must belong to an element in the fourth period. The only group 1 element in the fourth period is potassium.

The electron configuration of potassium is;

1s2 2s2 2p6 3s2 3p6 4s1

6 0

3 years ago