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lesya [120]
1 year ago
10

Transcribe and translate the following DNA sequence:

Chemistry
1 answer:
MA_775_DIABLO [31]1 year ago
7 0

Answer:

\\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \ \\  \\  \\  \\  \\ \ \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\

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10. Which of the following best describes the reaction 2VO3– (aq) + Zn (s) + 8H+ (aq) → 2VO2 (aq) + Zn2+ (aq) + 4H2O (l)? ______
Oduvanchick [21]
Correct answer is option E. <span>It is a redox reaction in which Zn is oxidized at the anode, and V is reduced at the cathode.

Reason:
In above reaction, the oxidation state of VO3- is +5, while that of VO2 is +4. Thus there is reduction of V from +5 to +4
In case of Zn, oxidation state of Zn is increased from 0 to +2, Thus process is referred as oxidation. </span>
3 0
3 years ago
A gaseous mixture contains 434.0 Torr H 2 ( g ) , 434.0 Torr H2(g), 389.9 Torr N 2 ( g ) , 389.9 Torr N2(g), and 77.9 Torr Ar (
weqwewe [10]

Answer:

H₂: 0.48,  N₂: 0.43;  Ar: 0.09

Explanation:

First of all, sum all the pressures to know the total pressure in the mixture.

434 Torr + 389.9 Torr + 77.9 Torr = 901.8 Torr

Mole fraction = Pressure gas / Total Pressure

Mole Fraction H₂: 434 Torr /901.8 Torr = 0.48

Mole Fraction N₂: 389.9 /901.8 Torr =0.43

Mole Fraction Ar:  77.9 /901.8 Torr = 0.09

Remember: <u>SUM OF MOLE FRACTION = 1</u>

8 0
3 years ago
Write the balanced oxidation-reduction reaction equation for the oxidation of benzoin by ammonium nitrate. 2
Margarita [4]
<span>Benzoin<span> is an organic compound with the formula PhCH(OH)C(O)Ph. It is a hydroxy ketone attached to two phenyl groups.</span><span>

To answer your question, </span><span>the balanced oxidation-reduction reaction equation for the oxidation of benzoin by ammonium nitrate is:

</span>2Ph-C(OH)-C(O)-Ph+NH4NO3 --> 2Ph-C(O)-C(O)-Ph + N2 + 3H2O.</span>

<span>
</span><span>I hope this helps and if you have any further questions, please don’t hesitate to ask again.</span>

7 0
3 years ago
The volume of a sphere is (4/3πr3 and the density of copper is 8.96g/cm3. part a what is the radius (in cm of a pure copper sphe
ira [324]
<span> <span> </span></span>Volume = 4/3 * PI * r^3

1.59 x 10^24 copper atoms = 2.64 moles of copper
Atomic Mass of copper = 63.55
2.64 * 63.55 = 167.77 grams of copper
Volume of Copper = Mass / Density
Volume of Copper = 167.77 grams / 8.96
Volume of Copper = <span> <span> 18.72</span></span> cubic centimeters

r^3 = Volume / (4/3 * PI)
r^3 = 18.72 / 4.188
r^3 = <span> <span> <span> 4.47
radius = </span></span></span><span><span><span>1.647</span> centimeters

</span></span>
8 0
3 years ago
The following (unbalanced) reaction is one of the steps to producing acid rain [as H2SO4(aq)] from sulfur-containing coal. In a
bulgar [2K]

Answer:

a) kc= [SO3 ]/([SO2 ][O2 ])

b) kc= 2.27*10⁶ M⁻¹

v) the reaction is product-favored

Explanation:

for the reaction, the equilibrium constant is

SO2 (g) + O2 (g) <-----> SO3 (g)

he equilibrum constant is

kc= [SO3 ]/([SO2 ]*[O2 ])

replacing values

kc= [SO3 ]/([SO2 ]*[O2 ]) = 1.01*10⁻² M/(3.61*10⁻³M*6.11 x 10⁻⁴ M) = 2.27*10⁶ M⁻¹

since kc>>1 the reaction is product-favored

4 0
3 years ago
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