Nucleotides have a phosphate group attached at the ___ carbon atom of the sugar.
1 answer:
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We have that the Complete Expanded Structure of (CH3)2CHCH2OCH2CH3 is given in the attachment below
From the Question
(CH3)2CHCH2OCH2CH3
Generally for the condensed formula (CH3)2CHCH2OCH2CH3
We consider that this is a single bond connecting them
We consider
Hydrogen H(1)
Oxygen(8)
Carbon(6)
In conclusion
The Complete Expanded Structure of (CH3)2CHCH2OCH2CH3 is given in the attachment below.
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Explanation:
It is known that the relation between pH and is as follows.
pH =
and,
Hence, first we will calculate the value of as follows.
=
= 4.75
Now, we will calculate the value of pH as follows.
pH =
=
= 4.75 + (-0.677)
= 4.07
Therefore, we can conclude that the pH of given solution is 4.07.
corrected question:
Determining Density and Using Density to Determine Volume or Mass
(a) Calculate the density of mercury if 1.00 × 10 g occupies a volume of 7.36 cm³
(b) Calculate the volume of 65.0 g of liquid methanol (wood alcohol) if its density is 0.791 g/mL.
(c) What is the mass in grams of a cube of gold (density = 19.32 g/cm) if the length of the cube is 2.00 cm?
(d) Calculate the density of a 374.5-g sample of copper if it has a volume of 41.8 cm³ A student needs 15.0 g of ethanol for an experiment. If the density of ethanol is 0.789 g/mL, how many milliliters of ethanol are needed? What is the mass, in grams, of 25.0 mL of mercury (density = 13.6 g/mL)?
Answer:
density =
ρ=m/v ,m=ρv, v=m/ρ
(a)m=1*10g , v=7.36cm³
ρ=10/7.36 =1.36g/cm³
(b) m=65g, ρ=0.791 g/mL.
v= 65/0.791 =82.17g/mL
(c) ρ=19.32g/cm³, l=2cm, v=l³=8cm³
m=19..32*8=154.56g/cm³
(d) mass of copper=374.5g , v=41.8cm³
ρ=374.5/41.8 =8.96g/cm³
mass of ethanol=15g, density of ethanol=0.789g/mL
v=15/0.789 =19.01mL
volume of mecury=25mL, density of mercury=13.6g/mL
m=25*13.6=340g