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3 years ago

A section of DNA has the base sequence GCTTAA. The corresponding messenger RNA base sequence will be

Chemistry

2 answers:

Viktor [21]3 years ago

6 0

G goes with c and t goes with a but the ts get changed to us so it would be CGAAUU.

Tems11 [23]3 years ago

4 0

It will be CGAAUU....

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Complete the reaction (NH4)2SO4(aq)+BaCl2(aq)

rosijanka [135]

(NH4)2SO4(aq) + BaCl2(aq) = BaSO4(s) + 2 NH4Cl(aq)
Reaction type: double replacement

5 0

4 years ago

For each given [H3O+] or [OH-], find the corresponding [OH-] or [H3O+] at 25°C.

-BARSIC- [3]

Answer:

a) [H3O+] = 1.00 E-10 M ⇒ [OH-] = 1.0 E-4 M

b) [H3O+] = 1.00 E-4 M ⇒ [OH-] = 1.0 E-10 M

c) [H3O+] = 9.90 E-6 M ⇒ [OH-] = 1.0 E-9 M

Explanation:

14 = pH + pOH
pH = - Log [H3O+]

a) [H3O+] = 1.00 E-10 M

⇒ pH = - Log(1.00 E-10) = 10

⇒ pOH = 14 - 10 = 4

⇒ 4 = - Log[OH-]

⇒ [OH-] = 1.0 E-4 M

b) [H3O+] = 1.00 E-4 M

⇒ pH = 4

⇒ pOH = 10

⇒ [OH-] = 1.0 E-10 M

c) [H3O+] = 9.90 E-6 M

⇒ pH = 5

⇒ pOH = 9

⇒ [OH-] = 1.0 E-9 M

8 0

3 years ago

Use Q = mcAT

mr Goodwill [35]

Answer:

1).....for the specific heat capacity(c) of water is 4200kg/J°C..

....guven mass(m)=320g(0.32kg)

...change in temperature(ΔT) =35°C

from the formula

Q=mcΔT

Q=0.32Kg x 4200kg/J°C x 35°C

Q=47,040Joules

5 0

3 years ago

Each continents were once joined together into one supercontinent what was it called Panterra, Panthalassa, Pangea, or Gia?

I am Lyosha [343]

Answer:

The correct answer is Pangea.

Explanation:

Pangea, also spelled Pangaea, in early geologic time, was a supercontinent that incorporated almost all the landmasses on Earth.

Hope this helps!

8 0

3 years ago

An industrial synthesis of urea obtains 87.5 kg of urea upon reaction of 68.2 kg of ammonia with excess carbon dioxide. Determin

dolphi86 [110]

Answer:

The theoretical yield of urea = 120.35kg

The percent yield for the reaction = 72.70%

Explanation:

Lets calculate -

The given reaction is -

$2NH_3(aq)+CO_2$ → $CH_4N_2O(aq)+H_2O (l)$

Molar mass of urea $CH_4N_2O$ = 60g/mole

Moles of $NH_3$ = $\frac{62.8kg/mole}{17g/mole}$ (since $Moles=\frac{mass of substance}{mass of one mole}$ )

= 4011.76 moles

Moles of $CO_2$ = $\frac{105kg}{44g/mole}$

= $\frac{105000g}{44g/mole}$

= 2386.36 moles

Theoritically , moles of $NH_3$ required = double the moles of $CO_2$

but , $4011.76$ , the limiting reagent is $NH_3$

Theoritical moles of urea obtained = $\frac{1 mole CH_4N_2O}{2mole NH_3}\times4011.76 mole NH_3$

= $2005.88mole CH_4N_2O$

Mass of 2005.88 mole of $CH_4N_2O$ = $2005.88 mole \times\frac{60g CH_4N_2O}{1mole CH_4N_2O}$

= 120352.8g

$120352.8g\times \frac{1kg}{1000g}$

= 120.35kg

Therefore , theroritical yeild of urea = 120.35kg

Now , Percent yeild = $\frac{87.5kg}{120.35kg}\times100$

72.70%

Thus , the percent yeild for the reaction is 72.70%

8 0

3 years ago