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CaHeK987 [17]
3 years ago
15

1.) Convert 202.1 km to millimeters

Chemistry
1 answer:
Dmitry_Shevchenko [17]3 years ago
3 0

Answer: 2.021 x 10^8 mm or 202,100,000

Explanation: Going from km to mm, you move the decimal place 6 times to the right.

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Select all of the statements that correctly describe DNA structure:
GalinKa [24]

Answer:

  1. Hydrogen bonds will hold two DNA strands together.
  2. The sugars and phosphate groups are located on the exterior of the helix.
  3. Base pairing will always be between a purine base and a pyrimidine base.

<em>I hope this helps you</em>

<em>:)</em>

3 0
2 years ago
When the pressure that a gas exerts on a sealed container changes from 22.5 psi to psi the temperature changes from 110 c to 65.
ad-work [718]
Gay- Lussacs law states that the pressure of a gas is directly proportional to temperature for a fixed amount of gas at constant volume.
therefore P/T = k
where P - pressure , T - temperature and k - constant 
\frac{P1}{T1} =  \frac{P2}{T2}&#10;
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation 
T1 - temperature in kelvin - 110 °C + 273 = 383 K
T2 - 65.0 °C + 273 = 338 K
substituting these values in the equation 

\frac{22.5 psi}{383 K} =  \frac{P}{338K}
P = 19.9 psi 
new pressure is 19.9 psi
3 0
3 years ago
The standard reduction potentials for the Ag+|Ag(s) and Zn2+| Zn(s) half-cell reactions are +0.799 V and -0.762 V, respectively.
mihalych1998 [28]

<u>Answer:</u> The potential of the given cell is 1.551 V

<u>Explanation:</u>

The given chemical cell follows:

Zn(s)|Zn^{2+}(0.125M)||Ag^{+}(0.240M)|Ag(s)

<u>Oxidation half reaction:</u> Zn(s)\rightarrow Zn^{2+}(0.125M)+2e^-;E^o_{Zn^{2+}/Zn}=-0.762V

<u>Reduction half reaction:</u> Ag^{+}(0.240M)+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.799V       ( × 2)

<u>Net cell reaction:</u> Zn(s)+2Ag^{+}(0.240M)\rightarrow Zn^{2+}(0.125M)+2Ag(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.799-(-0.762)=1.561V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Zn^{2+}]}{[Ag^{+}]^2}

where,

E_{cell} = electrode potential of the cell = ? V

E^o_{cell} = standard electrode potential of the cell = +1.561 V

n = number of electrons exchanged = 2

[Zn^{2+}]=0.125M

[Ag^{+}]=0.240M

Putting values in above equation, we get:

E_{cell}=1.561-\frac{0.059}{2}\times \log(\frac{(0.125)}{(0.240)^2})

E_{cell}=1.551V

Hence, the potential of the given cell is 1.551 V

6 0
4 years ago
An experiment with high validity is
uysha [10]

Answer:

extraneous variables.

Explanation:

5 0
3 years ago
Mary put a plant on a table in her room. After a few weeks, she saw that the stem of the plant was starting to bend toward the w
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 a. the need for light
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