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3 years ago

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Which structures perform similar functions in plant and animal cells?

1 answer:

lyudmila [28]3 years ago

8 0

Answer:

2

Explanation:

Sorry if Im wrong

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How many joules of heat are absorbed when 73 g water are heated from 30*C to 43*C? *

stepladder [879]

Answer:

3966.82 J

Explanation:

q=sm∆T

q=73×13×4.18

the specific heat for water is 4.18

6 0

3 years ago

Read 2 more answers

Which electron configuration represents a potassium atom in excited state

Anestetic [448]

Answer: The electronic configuration of potassium atom in excited state is $1s^22s^22p^63s^22p^64s^04p^1$

Explanation:

There are 2 states classified under energy levels:

1. Ground state: This is the lower energy state which is termed as stable state.

2. Excited state: This is the upper energy state and is termed as the unstable state. All the electrons which are present in this state always come back to the ground state.

Ground state electronic configuration of potassium (Z = 19) : $1s^22s^22p^63s^22p^64s^1$

As, only 1 electron can freely move to upper state. Hence, the excited state electronic configuration of potassium will be: $1s^22s^22p^63s^22p^64s^04p^1$

5 0

3 years ago

Read 2 more answers

The normal freezing point of water (H2O) is 0.00 oC and its Kf value is 1.86 oC/m. A nonvolatile, nonelectrolyte that dissolves

adelina 88 [10]

Answer:

3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C

Explanation:

One colligative property is the freezing point depression due the addition of a solute. The equation is:

ΔT=Kf*m*i

<em>Where ΔT is change in temperature = 0.400°C</em>

<em>Kf is freezing point constant of the solvent = 1.86°C/m</em>

<em>m is molality of the solution (Moles of solute / kg of solvent)</em>

<em>And i is Van't Hoff constant (1 for a nonelectrolyte)</em>

Replacing:

0.400°C =1.86°C/m*m*1

0.400°C / 1.86°C/m*1 = 0.215m

As mass of solvent is 280.0g = 0.2800kg, the moles of the solute are:

0.2800kg * (0.215moles / 1kg) = 0.0602 moles of solute must be added.

The mass of ethylene glycol must be added is:

0.0602 moles * (62.10g / mol) =

3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C

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6 0

3 years ago

Four people make observations recorded in the table.

77julia77 [94]

Answer:

D. They involve changes in energy.

The last one

Explanation:

have a good day

6 0

2 years ago

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The total mass of the atmosphere is about 5.00 x 1018 kg. How many moles each of air, O2, and CO2 are present in the atmosphere?

n200080 [17]

<u>Answer:</u> The moles of oxygen and carbon dioxide in air is $3.63\times 10^{19}mol$ and $7.18\times 10^{16}mol$ respectively

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of atmosphere = $5.00\times 10^{18}kg=5.00\times 10^{21}g$

Average molar mass of atmosphere = 28.96 g/mol

Putting values in above equation, we get:

$\text{Moles of atmosphere}=\frac{5.00\times 10^{21}g}{28.96g/mol}=1.73\times 10^{20}mol$

We know that:

Percent of oxygen in air = 21 %

Percent of carbon dioxide in air = 0.0415 %

Moles of oxygen in air = $\frac{21}{100}\times 1.73\times 10^{20}=3.63\times 10^{19}mol$

Moles of carbon dioxide in air = $\frac{0.0415}{100}\times 1.73\times 10^{20}=7.18\times 10^{16}mol$

Hence, the moles of oxygen and carbon dioxide in air is $3.63\times 10^{19}mol$ and $7.18\times 10^{16}mol$ respectively

6 0

3 years ago