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vesna_86 [32]
3 years ago
10

Draw the structures of organic compounds a and b. omit all byproducts

Chemistry
1 answer:
Georgia [21]3 years ago
8 0

81. There is 1 carbon, 2 chlorine and fluorine atoms in Freon 12. To draw them it forms a cross with C in the middle and Cl and F both on the opposite side. 
     


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Explain how you would separate a mixture of sand and water.
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Answer: When sand is added to water it either hangs in the water or forms a layer at the bottom of the container. Sand therefore does not dissolve in water and is insoluble. It is easy to separate sand and water by filtering the mixture. Salt can be separated from a solution through evaporation.

Explanation:

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Need help ASAP with this question
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What kinds of substances besides water can be involved in hydrogen bonding?
MrMuchimi
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The best example of hydrogen bonding excluding water is DNA. The two strands of polymers are connected by hydrogen bonds between the nucleotide bases</span>.
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Write the balanced chemical equation for the dissociation reaction of thiosulfuric acid.
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A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base
tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ =  0.00375  mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

HN_3 + OH- ---> N^-_{3} + H_2O

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction =  0.025 +  0.0133 = 0.0383  L

Concentration of HN_3 = \dfrac{0.001755}{0.0383} = 0.0458 M

Concentration of N^{-}_3 = \dfrac{ 0.001995 }{0.0383} = 0.0521 M

GIven that :

Ka = 1.9 x 10^{-5}

Thus; it's pKa = 4.72

pH =4.72 +  log(\dfrac{ \ 0.0521}{0.0458})

pH =4.72 + log(1.1376)

pH =4.72 + 0.05598

pH =4.77598

pH ≅ 4.80

3 0
3 years ago
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