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4 years ago

Explain how cobalt-60 and cobalt -59 can be the same element but have different mass numbers

1 answer:

vladimir1956 [14]4 years ago

3 0

The number of protons and the atomic number are the same for both forms of cobalt, but the number of neutrons varies for each cobalt ion.

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I need help solving this chemistry question

Arturiano [62]

Answer:

I think the answer is 22.2

Explanation: What i DID was adding 12.0 + 10.0 and than that gave me 22 the I had added the to 0.200 and that how i got 22.2. Sorry if i got is wrong. :(

3 0

4 years ago

How are they related "atom" "molecule"

zhenek [66]

Answer:

mixture of atoms forms molecule

7 0

3 years ago

Read 2 more answers

What is bias in an experiment?

N76 [4]

○ c. ○

◘

because a Research bias, also called experimenter bias, is a process where the scientists performing the research influence the results, in order to portray a certain outcome.

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hope that helped ♥

3 0

3 years ago

Partner Namets) For each of the following reactions carried out: Write the balanced chemical equation, the full ionic equation,

torisob [31]

Answer : The full balanced ionic equation will be,

$2KI(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbI_2(s)$

Reactants are lead nitrate and potassium iodide.

Products are lead iodide and potassium nitrate.

The spectator ions are, $K^+,NO_3^-$

Explanation :

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

When potassium iodide react with lead nitrate then it gives potassium nitrate and lead iodide as a product.

The full balanced ionic equation will be,

$2KI(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbI_2(s)$

The ionic equation in separated aqueous solution will be,

$2K^+(aq)+2I^{-}(aq)+Pb^{2+}(aq)+2NO_3^{-}(aq)\rightarrow PbI_2(s)+2K^+(aq)+2NO_3^{-}(aq)$

In this equation, $K^+\text{ and }NO_3^-$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$Pb^{2+}(aq)+2I^{-}(aq)\rightarrow PbI_2(s)$

4 0

4 years ago

The Kp for the reaction below is 1.49 × 108 at 100.0°C:

Anna007 [38]

Kp= (COCl2)/[(CO)(Cl2)]= 1.49 x 10^8

1.49 x 10^8= (COCl2/((2.22x10-4)(2.22x10-4))

COCl2= 1.49x10^8 x ((2.22x10-4)(2.22x10-4))= 7.34 atm

5 0

3 years ago