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Naya [18.7K]
2 years ago
14

I have five less protons than the least massive metalloid in the fourth period. Who am I?

Chemistry
1 answer:
exis [7]2 years ago
6 0

Answer:

You are the Cobalt

Explanation:

The least massive metalloid in the fourth period is Germanium, and it have 32 protons. If you have 5 less protons: 32 - 5 = 27 protons. The element with 27 protons is Cobalt

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VARVARA [1.3K]

Answer:

I think the answer is A

3 0
3 years ago
Identify the oxidizing and reducing agents in the following: H2S(aq) + Cl2(g) -> S(s) + 2HCI (aq)
alexgriva [62]

The oxidizing and reducing agent in the above redox reaction are hydrogen sulphide (H2S) and Chlorine (Cl) respectively.

<h3>What is an oxidizing and reducing agent?</h3>

An oxidizing agent is any substance that oxidizes, or receives electrons from another substance and as a result, becoming reduced.

On the other hand, a reducing agent is any substance that reduces or donates electrons to another and as a result becomes oxidized.

According to this reaction; H2S(aq) + Cl2(g) -> S(s) + 2HCI (aq)

  • H2S accepts electrons from Cl2 and becomes reduced to S
  • Cl2 donates electrons to H2S and becomes oxidized to HCl

Therefore, the oxidizing and reducing agent in the above redox reaction are hydrogen sulphide (H2S) and Chlorine (Cl) respectively.

Learn more about oxidizing agent at: brainly.com/question/10547418

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8 0
1 year ago
Consider a certain type of nucleus that has a half-life of 32 min. calculate the percent of original sample of nuclides remainin
Irina18 [472]
t1/2 = ln 2 / λ = 0.693 / λ
Where t1/2 is the half life of the element and λ is decay constant.

32 = 0.693 / λ 
λ   = 0.693 / 32          (1) 

Nt = Nο eΛ(-λt)          (2)

Where Nt is atoms at t time, λ is decay constant and t is the time taken.
t = 1.9 hours = 1.9 x 60 min

From (1) and (2),


Nt = Nο e⁻Λ(0.693/32)*1.9*60
Nt =  0.085Nο 

Percentage = (Nt/Nο) x 100%
                   = (0.085Nο/Nο) x 100%
                   = 8.5%

Hence, Percentage of remaining atoms with the original sample is 8.5%

8 0
3 years ago
Given a 0.200 M solution of anserine at its isoelectric point and ready access to 0.100 M HCl, 0.100 M NaOH and distilled water,
vekshin1

Solution :

Given :

Amount of anserine solution = 0.200 M

pH value is = 7.20

Preparation of 0.04 M solution of anserine from the 0.2 M solution.

0.2 M x x = 0.04 M x 1000 ml

x = 200 ml

So the 200 ml of 0.2 M anserine solution is required to prepare0.04 M of anserine.

0.1 M x x = 0.04 x 1000 ml

x = 400 ml

Therefore, 400 ml of HCl is needed.

6 0
2 years ago
What happens to light when it is captured by a material?<br> plz, help meeeee!!!!! TwT
Nat2105 [25]
During absorption, frequency of the incoming light wave is either near or at the energy level of the electrons in the matter of the material. The electrons will absorb the energy of the light wave and change their energy state.

Hope that helps! :)
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