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blagie [28]
3 years ago
9

Which is a single ringed nitrogeo base? A)adenine B)guanine C)thymine D)ribose

Chemistry
1 answer:
pogonyaev3 years ago
3 0
A single ringed nitrogeo base is thymine (C)

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Plz help me with this question, I am stuck
zhenek [66]

i only know that E is one of them

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3 years ago
Which process would produce a similar pathway? condensation dehydration decomposition melting
babymother [125]
It is A) Condensation
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3 years ago
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Benzene and toluene form ideal solutions. Consider a solution of benzene and toluene prepared at 25∘C. As suming that the mole f
Montano1993 [528]

Answer:

XL(toluene) = 0.77

XL(benzene) = 0.23

Explanation: see attachment

7 0
2 years ago
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A slurry of flakes soybeans weighing a total of 100 kg contains 75 kg of inert solids and 25 kg of solution with 10 wt% oil and
lubasha [3.4K]

Answer:

the amounts and compositions of the overflow V1 and underflow L1 leaving the stage are 75kg and 125kg respectively.

Explanation:

Let state the given parameters;

Let A= solvent (hexane)

B= solid(inert soiid)

C= solvent(oil)

F_{solution} = mass of solvent + mass of oil (i.e A+C)

<u>Feed Phase:</u>

Total feed (i.e slurry of flakes soybeans)= 100kg

B= mass of solid =75 kg

F= mass of solvent + mass of oil (i.e A+C)

 = 25kg

Mass ratio of oil to solution Y_{F} =\frac{Mass C}{Mass (A+C)}

mass of oil (C) =25 × 0.1 wt = 2.5kg

mass of hexane  in feed = 25 ×  0.9 =22.5kg + 2.5 =25kg

therefore  Y_{F} = \frac{2.5}{25}

= 0.1

mass ratio of solid to solution Y_{A}  =  \frac{Mass A}{Mass (A+C)}=[tex]\frac{75}{25}

=3

<u>Solvent Phase:</u>

C= Mass of oil= 0(kg)

A= Mass of hexane = 100kg

mass of solutions = A+C = 0+100kg

solvent= 100kg

<u>Underflow:</u>

underflow = L₁ = (unknown) ???

L₁ = E₁ + B

the value of N for the outlet and underflow is 1.5 kg

i.e N₁ = \frac{mass B}{mass(A+C)}

solution in underflow E₁ = Mass (A+C)

<u>Overflow:</u>

Overflow = V₁ = (unknown) ???

solution in overflow V₁ = Mass (A+C)

This is because, B = 0 in overflow

Solid Balance: (since the solid is inert, then is said to be same in feed & underflow).

solid in feed = solid in underflow = 75

75=  E₁ × N₁

75 =  E₁ × 1.5

E₁ = 50kg

Underflow L₁ = E₁ × B

= 50 + 75

=125kg

The Overall Balance: Feed + Solvent = underflow + overflow

100 + 100 = 125 + V₁

V₁ = 75kg

5 0
3 years ago
What will be the volume occupied by 2.5 moles of nitrogen gas exerting 1.75 atm of pressure at 475K?
Marina86 [1]

Answer:

THE VOLUME OF THE NITROGEN GAS AT 2.5  MOLES , 1.75 ATM AND 475 K IS 55.64 L

Explanation:

Using the ideal gas equation

PV = nRT

P = 1.75 atm

n = 2.5 moles

T = 475 K

R = 0.082 L atm/mol K

V = unknown

Substituting the variables into the equation we have:

V = nRT / P

V = 2.5 * 0.082 * 475 / 1.75

V = 97.375 / 1.75

V = 55.64 L

The volume of the 2.5 moles of nitrogen gas exerted by 1.75 atm at 475 K is 55.64 L

6 0
2 years ago
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