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AysviL [449]
2 years ago
12

The________ cation will precipitate after the_________ step. Then the _________cation will precipitate after the________ step.

Chemistry
1 answer:
Novay_Z [31]2 years ago
8 0

Answer:

f. Sn^4+

c. second

e. Al^3+

d. third

Explanation:

This question comes from a quantitative analysis showing the flowchart of a common scheme for identifying cations.

Now, from the separation scheme, Let's assume that Sn⁴⁺ & Al³⁺ were given; Then, Yes, the separation will work.

However, there will be occurrence of precipitation after the 1st step1.

So,  the <u>Sn⁴⁺</u> cation will precipitate after the <u>second </u>step. Then the <u>Al³⁺</u> cation will precipitate after the <u>third</u> step.

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6. What are two uses for nonmetals?
lesantik [10]
Uses of nonmetals in our daily life: Oxygen which is 21% by volume helps in the respiration process.

Nonmetals used in fertilizers: Fertilizers contain nitrogen.

Nonmetals used in crackers: Sulphur and phosphorus are used in fireworks.
5 0
2 years ago
Why is it important to make observations about the reactants? ​
Free_Kalibri [48]

Answer:

Reason Down below

Explanation:

It is important because when you make observation you get a clue sometimes and it  reactants i feel like it also takes places with observation. :)

7 0
3 years ago
Which surface ocean currents has the warmest water
NNADVOKAT [17]
<span>the Brazil Current has the warmest water</span>
4 0
2 years ago
Read 2 more answers
What is the coefficient for hydrogen in the balanced equation for the reaction of solid molybdenum(iv) oxide with gaseous hydrog
anastassius [24]

The  coefficient  for hydrogen  in  the  balanced  equation   of solid  molybdenum(iV)  oxide  with gaseous  hydrogen  is  2


 Explanation

Coefficient   is defined  to  as  a number  in front   of a chemical formula in a  balanced chemical equation.

   The  reaction   of   molybdenum (iv) oxide  with  gaseous  hydrogen  is  as below,


MoO2  + 2 H2→  Mo  +2 H2O

From   balanced equation above  the coefficient   for  H2  is  2  since  the number in  front of  H2  is 2



3 0
3 years ago
2 CuCl2 + 4 KI → 2 CuI + 4 KCl + I2
Margaret [11]

Answer: 6.75 moles

Explanation:

This is a simple stoichiometry proboe. that I would set up like this:

(13.5 moles CuCI2) (1 mol I2 / 2 moles CuCi2)

That means you all you have to do for this problem is divide by 2 and cancel out the unit moles CuCI2, which leaves you with 6.75 moles I2.

Hope this helps :)

7 0
2 years ago
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