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meriva
2 years ago
10

An atom of an element that has the same number of protons but a different number of neutrons is called a(n): ion nucleus isotope

mixture
Chemistry
1 answer:
alexandr1967 [171]2 years ago
5 0
It is called an isotope
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what is the oxidation state of each element in coh2? c o h what is the oxidation state of each element in febr3? fe br
fgiga [73]

The oxidation state of the elements  in the compounds are:

CoH₂:

  • Co = +2
  • H = -1

FeBr₃:

  • Fe = +3
  • Br = -1

<h3>What is the oxidation states of the elements in the given compounds?</h3>

The oxidation states of the elements in each of the given compounds is determined as follows:

Cobalt dihydride, CoH₂

Co = +2

H = -1

Iron (iii) bromide, FeBr₃

Fe = +3

Br = -1

In conclusion, the oxidation state of the elements are charges they have in the compound.

Learn more about oxidation state at: brainly.com/question/27239694

#SPJ1

8 0
1 year ago
You probably already know the names of some molecules. Identify which compound (water, salt, carbon dioxide, or methane) corresp
vladimir1956 [14]

CO2 is carbon

H2O is water

CH4 is methane

NaCI is salt

7 0
3 years ago
Read 2 more answers
A part of the periodic table is shown below:
mylen [45]
I think is oxygen that what i think
8 0
3 years ago
Read 2 more answers
How many moles are in 1.51 x 10^24 molecules of water?
lyudmila [28]

Answer:One mole of HBr has 6.02 x 1

0

23

molecules of HBr.

1 mole of HBr = 6.02 x 1

0

23

molecules of HBr.-----(a)

X mole of HBr has 1.21 x

10

24

molecules of HBr.

X mole of HBr = 1.21 x

10

24

molecules of HBr------(b)

Taking ratio of (a) and (b)

X / 1 = 1.21 x

10

24

/ 6.02 x 1

0

23

X= 2.009 moles.

Explanation:

3 0
3 years ago
How much heat energy is required to raise the temperature of 0.360 kg of copper from 23.0 ∘C to 60.0 ∘C? The specific heat of co
AVprozaik [17]
MThe  heat  energy  required  to  raise  the  temperature   of  0.36Kg   of  copper   from   22 c   to  60  c  is  calculate  using  the  following  formula

MC delta T
m(mass)=  0.360kg  in  grams  =  0.360  x1000 = 360 g
  c(specific  heat  energy)  =  0.0920  cal/g.c
delta T =  60- 23  = 37  c

heat  energy is therefore=  360g   x0.0920 cal/g.c  x 37  c=  1225.44  cal

5 0
3 years ago
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