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NemiM [27]
2 years ago
11

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

Chemistry
2 answers:
Len [333]2 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]2 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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\qquad\qquad\huge\underline{{\sf Answer}}

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1 mole of \sf{ COCl_2} has 6.022 × 10²³ molecules of the given compound.

So, 0.78 mole of \sf{COCl_2} will have ~

\qquad \sf  \dashrightarrow \: 0.78 \times 6.022 \times 10 {}^{23}

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An unknown compound displays singlets at δ 2.1 ppm and 2.56 ppm in the ratio of 3:2. what is the structure of the compound?
olasank [31]

1) As can be seen from any 1H NMR chemical shift ppm tables, hydrogens which have δ values from 2ppm to 2.3ppm are hydrogens from carbon which is bonded to a carbonyl group. From this, we can conclude that our hydrogens belong to the type, but from 2 different alkyl groups because of 2 different signals.

 

2) So, one alkyl group is CH3 and second one can be CH or CH2.

 

3) If we know that ratio between two types of hydrogens is 3:2, it can be concluded that second alkyl group is CH2. 


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50.0 ml of 0.010m naoh was titrated with 0.50m hcl using a dropper pipet. if the average drop from the pipet has a volume of 0.0
creativ13 [48]

25 drops of acid is required to neutralize the 50.0 ml of 0.010m of NaOH in the experiment.

The equation of the reaction is;

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We can use the titration formula;

CAVA/CBVB = NA/NB

CA= concentration of acid

VA = volume of acid

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NB = 1

Substituting values;

CAVANB = CBVBNA

VA =  0.010 ×  50.0 × 1/ 0.50 × 1

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Since the total volume of acid used is 1 ml and each drop contains 0.040 ml

The number of drops required is 1ml/0.040 ml = 25 drops

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