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Afina-wow [57]
3 years ago
9

The mass of a sample of sodium bicarbonate is 2.1 kilograms (kg). There are 1,000 grams (g) in 1 kg, and 1 mc020-1.jpg 109 nanog

rams (ng) in 1 g. What is the mass of this sample in ng? 2.1 mc020-2.jpg 103 ng 2.1 mc020-3.jpg 106 ng 2.1 mc020-4.jpg 109 ng 2.1 mc020-5.jpg 1012 ng
Chemistry
2 answers:
zalisa [80]3 years ago
4 0

Answer:The mass of sodium bicarbonate in nano grams is 1.2\times 10^{12} nano grams.

Explanation:

Given:

1000 g = 1kg

10^9 ng=1 g

Mass of the sodium bicarbonate = 2.1 kg

Mass of the sodium bicarbonate in grams =2.1 kg\times 1000 g=2100 g

Mass of the sodium bicarbonate in ng = 2100\times 10^9 ng = 2.1\times 10^{12} ng

The mass of sodium bicarbonate in nano grams is 1.2\times 10^{12} nano grams.

gulaghasi [49]3 years ago
3 0

Answer: The mass of the sample having 2.1 kg of NaHCO_3 will be 2.1\times 10^{12}ng

Explanation: We are given a sample of sodium bicarbonate (NaHCO_3) having mass 2.1 kg

To convert this into nano-grams, we use the following conversions:

1 kg = 1000 grams

and

1g=10^9nanograms

Converting 2.1 kg to nano-grams using above conversions:

2.1kg=2.1\times (1000\times 10^9)ng=2.1\times 10^{12}ng

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Answer:

Explanation:

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   III                                +3                                                          3

   IV                           +4 or -4                                                     4

   V                                  -3                                                         5

   VI                                 -2                                                         6

  VII                                 -1                                                          7

  VIII                                 0                                                          8

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3 years ago
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2. In cyclohexane molecule as the molecule adopts a chair conformation in order  to eliminate the torsional strain which would occur if the cyclohexane ring were to be planar. Torsional strain is basically the inter electronic repulsion between the atoms that do not share a bond. So this strain happens on account of eclipsing atoms. In case of eclipse structure there would be a lot of torsional strain. In case of chair conformation all the C-H bonds happen to be completely staggered in nature to eliminate the torsional strain.

3. The ring strain in case of cycloalkanes are  dependent upon the number of CH₂ groups present as that would determine the size of the ring and subsequently its structure ,whether the ring would be 5 , 6 or 7 membered  .Cyclohexane is a 6 -membered as there are 6CH₂ groups in it and the existence of chair conformation is only for Cyclohexane or for molecules having 6-membered ring  . Any change  in number of CH₂groups would lead to a different conformational structure.

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6.All the bonds in cyclohexane ring are staggered to eliminate the torsional strain. It is quite evident that the cyclohexane ring is completely stable free of the ring strain.So there are no eclipsing bonds present in cyclohexane.

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