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Nookie1986 [14]
2 years ago
14

in a petri dish, a certain type of bacterium doubles in number every 30 minutes. there were originally 16, or 2^4, bacteria in t

he dish. after 150 minutes, the number of bacteria has doubled 7 times, multiplying by 2^7. Now the population of bacteria is 2^4 • 2^7. Expressed as a power, how many bacteria are in the petri dish after 150 minutes? A. 2^28 B. 2^11 C.4^11 D.2^7
Chemistry
1 answer:
mafiozo [28]2 years ago
6 0

Answer:

D

Explanation:

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3 years ago
Quiz due soon need help ASAP
Rom4ik [11]

Answer:

3

Explanation:

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2 years ago
A 455 g piece of bopper tubing is heated to 89.5 C and placed in an insulated vessel containing 159 g of water at 22.8 C. Assumi
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Answer:

The final temperature of the setup = 36.6°C

Explanation:

Let the final temperature of the setup be T

Heat lost by the copper tubing = Heat gained by water and the vessel

Heat lost by the copper tubing = mC ΔT = 455 × 0.387 × (89.5 - T) = (15759.61 - 176.1T) J

Heat gained by water = mC ΔT = 159 × 4.186 × (T - 22.8) = (665.6T - 15175.1) J

Heat gained by vessel = c ΔT = 10 × (T - 22.8) = (10T - 228) J

Heat lost by the copper tubing = Heat gained by water and the vessel

(15759.61 - 176.1T) = (665.6T - 15175.1) + (10T - 228)

851.7 T = 31162.71

T = 36.6°C

7 0
2 years ago
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kvasek [131]
B I think
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2 years ago
Carbon disulfide has an unusual property of being able to dissolve several nonmetals. What are the boiling points and freezing p
nydimaria [60]

<u>Answer:</u> The freezing and boiling points of the solution is -114.34°C and 48.95°C respectively

<u>Explanation:</u>

  • <u>Calculating the freezing point:</u>

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = -110.8°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point constant = 3.83°C/m

m_{solute} = Given mass of solute (P_4) = 8.44 g

M_{solute} = Molar mass of solute (P_4) = 124 g/mol

W_{solvent} = Mass of solvent (carbon disulfide) = 60.0 g

Putting values in above equation, we get:

-110.8-\text{Freezing point of solution}=1\times 3.83^oC/m\times \frac{8.44\times 1000}{124g/mol\times 60.0}\\\\\text{Freezing point of solution}=-114.34^oC

Hence, the freezing point of solution is -114.34°C

  • <u>Calculating the boiling point:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

Or,

\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Boiling point of pure solution = 46.3°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_b = molal boiling point constant = 2.34°C/m

m_{solute} = Given mass of solute (P_4) = 8.44 g

M_{solute} = Molar mass of solute (P_4) = 124 g/mol

W_{solvent} = Mass of solvent (carbon disulfide) = 60.0 g

Putting values in above equation, we get:

\text{Boiling point of solution}-46.3=1\times 2.34^oC/m\times \frac{8.44\times 1000}{124g/mol\times 60.0}\\\\\text{Boiling point of solution}=48.95^oC

Hence, the boiling point of solution is 48.95°C

7 0
2 years ago
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