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matrenka [14]
2 years ago
15

Approximately ____ days have passed when 20% of Cobalt (Co) remains? A) 19 B) 20 C) 21 D) 29

Chemistry
1 answer:
Brums [2.3K]2 years ago
3 0
The answer to this is A i think.
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The student decided to do another experiment with his leftover copper(II) sulfate (CuSO4) solution. He divided the solution up i
lakkis [162]

Answer:

CuSO4 + Fe -> FeSO4 + Cu

Explanation:

This reaction is a classic example of a redox reaction. I won't go in too deep, but the basic thing is that electrons from the Fe atom go to the Cu2+ ion. Therefore, Fe becomes an ion, and Cu - an electroneutral atom:

Fe + Cu2+ -> Fe2+ + Cu.

Silver is not a very reactive metal and it does not give up its electrons to Cu.

6 0
2 years ago
A gas balloon has a volume of 80.0 mL at 300K , and a pressure of 50.0 kPa. if the pressure changes to 80.0 kPa and the temperat
stellarik [79]

Answer: 53.3

Explanation:

V2=(T2 x P1 x V1)/(T1 x P2)

(320x50x80)/(300x80)

53.3

3 0
2 years ago
What is the oxidation number for iodine in Mg(IO3)2 ?
navik [9.2K]
The oxidation number of iodine is 5 in Mg(IO3)2 which can be calculated as 
   Mg(IO3)2
   MgI2O6
As we know that
Mg has +2
O has -2
So,
   (+2) + 2I + 6 (-2)=0
   2 + 2I - 12 =0
   10+ 2I =0
    10 = 2I
     I =5

7 0
2 years ago
Read 2 more answers
Luis is helping his parents paint a border around the walls of a room. He uses a stencil to repeat the same design on each wall
Vinvika [58]
The answer to your question is d
3 2
2 years ago
Read 3 more answers
How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation? 2HBr(aq)+
Mkey [24]

Full Question:

A flask containing 420 Ml of 0.450 M HBr was accidentally knocked to the floor.?

How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation?

2HBr(aq)+K2CO3(aq) ---> 2KBr(aq) + CO1(g) + H2O(l)

Answer:

13.1 g K2CO3 required to neutralize spill

Explanation:

2HBr(aq) + K2CO3(aq) → 2KBr(aq) + CO2(g) + H2O(l)

Number of moles = Volume * Molar Concentration

moles HBr= 0.42L x .45 M= 0.189 moles HBr

From the stoichiometry of the reaction;

1 mole of K2CO3 reacts  with 2 moles of HBr

1 mole = 2 mole

x mole = 0.189

x = 0.189 / 2 = 0.0945 moles

Mass = Number of moles * Molar mass

Mass = 0.0945 * 138.21  = 13.1 g

3 0
2 years ago
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