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LenKa [72]
3 years ago
10

Please explain this to me, it's taking forever for me!

Chemistry
1 answer:
liubo4ka [24]3 years ago
5 0

Answer:

civic cgvvbbbibbbnn bnnnn

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The weight of the body in the air is .... the weight of the submerged body
stira [4]

Answer:

the correct answer is option 'b': More than

Explanation:

The 2 situations are represented in the attached figures below

When an object is placed in air it is acted upon by force of gravity of earth which is measured as weight of the object.

While as when any object is submerged partially or completely in any fluid the fluid exerts a force in upward direction and this force is known as force of buoyancy and it's magnitude is given by Archimedes law as equal to the weight of the fluid that the body displaces, hence the effective force in the downward direction direction thus the apparent weight of the object in water decreases.

4 0
3 years ago
I need help with a questino
kipiarov [429]

Explanation:

What will the question be ?

6 0
3 years ago
Read 2 more answers
How many particles are in one mole of copper (II) sulfate, CuSO4?
Kobotan [32]
2. Avogadro’s number
3 0
3 years ago
An iron ore sample weighing 0.5562 g is dissolved HCl (aq), and the iron is obtained as Fe2 in solution. This solution is then t
erik [133]

Answer:

\% Fe^{+2}=70%

Explanation:

Hello,

In this case, we could considering this as a redox titration:

Fe^{+2}+(Cr_2O_7)^{-2}\rightarrow Fe^{+3}+Cr^{+3}

Thus, the balance turns out (by adding both hydrogen ions and water):

Fe^{+2}\rightarrow Fe^{+3}+1e^-\\(Cr_2^{+6}O_7)^{-2}+14H^++6e^-\rightarrow 2Cr^{+3}+7H_2O\\\\6Fe^{+2}+(Cr_2^{+6}O_7)^{-2}+14H^++6e^-\rightarrow Cr^{+3}+7H_2O+6Fe^{+3}+6e^-

Thus, by stoichiometry, the grams of Fe+2 ions result:

m_{Fe^{+2}}=0.04021\frac{molK_2Cr_2O_7}{L}*0.02872L*\frac{6molFe^{+2}}{1molK_2Cr_2O_7}*\frac{56gFe^{+2}}{1molFe^{+2}}=0.388gFe^{+2}

Finally, the mass percent is:

\% Fe^{+2}=\frac{0.388g}{0.5562g}*100\%\\  \% Fe^{+2}=70%

Best regards.

8 0
3 years ago
What volume of H2O is formed at stp when 6.0g of Al is treated with excess NaOH? NaOH + Al + H2O —-> NaAl(OH)4 + H2 (g)
ioda

this is the formula and answer of this question

8 0
3 years ago
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