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NemiM [27]
3 years ago
11

Which of the following is not the same as 2.97 milligrams?

Chemistry
2 answers:
Len [333]3 years ago
5 0
Hi There!!!!!!

Which of the following is not the same as 2.97 milligrams?A)0.297 cg
B)0.00297 g
C) 0.0000297 KG

Answer:0.00297 g

Evgesh-ka [11]3 years ago
3 0

Answer : The (C) 0.0000297 kg is not the same as 2.97 mg.

Explanation :

The conversion used for mass from milligram to centigram is:

1mg=10^{-1}cg

The conversion used for mass from milligram to gram is:

1mg=10^{-3}g

The conversion used for mass from milligram to kilogram is:

1mg=10^{-6}kg

As we are given that the mass 2.97 mg. Now we have to determine the mass in cg, g and kg.

Mass of 'cg' :

1mg=10^{-1}cg

2.97mg=\frac{2.97mg}{1mg}\times 10^{-1}cg=2.97\times 10^{-1}cg=0.297cg

Mass of 'g' :

1mg=10^{-3}g

2.97mg=\frac{2.97mg}{1mg}\times 10^{-3}g=2.97\times 10^{-3}g=0.00297g

Mass of 'kg' :

1mg=10^{-6}kg

2.97mg=\frac{2.97mg}{1mg}\times 10^{-6}kg=2.97\times 10^{-6}kg=0.00000297kg

Hence, from this we conclude that, 0.0000297 kg is not the same as 2.97 mg.

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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
A 100.0 ml sample of 0.10 m ca(oh)2 is titrated with 0.10 m hbr. determine the ph of the solution after the addition of 400.0 ml
WINSTONCH [101]
The balanced equation for the reaction is as follows;
Ca(OH)₂ + 2HBr --> CaBr₂ + 2H₂O
stoichiometry of Ca(OH)₂ to HBr is 1:2
number of Ca(OH)₂ moles reacted - 0.10 mol/L x 0.1000 L = 0.010 mol
Number of HBr moles added - 0.10 mol/L x 0.4000 = 0.040 mol 
1 mol of Ca(OH)₂ needs 2 mol of HBr for neutralisation
therefore 0.010 mol of Ca(OH)₂  needs - 0.010 x 2 = 0.020 mol of HBr to be neutralised
but 0.040 mol of HBr has been added therefore number of moles of HBr in excess - 0.040 - 0.020 = 0.020 mol 
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HBr is a strong acid therefore complete ionization
[HBr] = [H⁺]
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pH = -log[H⁺] 
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pH = 1.40
pH of the medium is 1.40
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