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3 years ago

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At midnight in the dark of the night, what is most likely happening in a leaf? * O Respiration O Photosynthesis O Both photosynt

hesis and respiration Neither photosynthesis or respiration
Please help thanks!

2 answers:

kumpel [21]3 years ago

5 0

Answer:2

Explanationb

sammy [17]3 years ago

4 0

Answer: Photosynthesis

Explanation:

You might be interested in

At a certain temperature the vapor pressure of pure benzene is measured to be . Suppose a solution is prepared by mixing of benz

Marianna [84]

Answer:

P(C₆H₆) = 0.2961 atm

Explanation:

I found an exercise pretty similar to this, so i'm gonna use the data of this exercise to show you how to do it, and then, replace your data in the procedure so you can have an accurate result:

<em>"At a certain temperature the vapor pressure of pure benzene (C6H6) is measured to be 0.63 atm. Suppose a solution is prepared by mixing 79.2 g of benzene and 115. g of heptane (C7H16) Calculate the partial pressure of benzene vapor above this solution. Round your answer to 2 significant digits. Note for advanced students: you may assume the solution is ideal".</em>

<em />

Now, according to the data, we want partial pressure of benzene, so we need to use Raoul's law which is:

P = Xₐ * P° (1)

Where:

P: Partial pressure

Xₐ: molar fraction

P°: Vapour pressure

We only have the vapour pressure of benzene in the mixture. We need to determine the molar fraction first. To do this, we need the moles of each compound in the mixture.

To get the moles: n = m / MM

To get the molar mass of benzene (C₆H₆) and heptane (C₇H₁₆), we need the atomic weights of Carbon and hydrogen, which are 12 g/mol and 1 g/mol:

MM(C₆H₆) = (12*6) + (6*1) = 78 g/mol

MM(C₇H₁₆) = (7*12) + (16*1) = 100 g/mol

Let's determine the moles of each compound:

moles (C₆H₆) = 79.2 / 78 = 1.02 moles

moles (C₇H₁₆) = 115 / 100 = 1.15 moles

moles in solution = 1.02 + 1.15 = 2.17 moles

To get the molar fractions, we use the following expression:

Xₐ = moles(C₆H₆) / moles in solution

Xₐ = 1.02 / 2.17 = 0.47

Finally, the partial pressure is:

P(C₆H₆) = 0.47 * 0.63

<h2>P(C₆H₆) = 0.2961 atm</h2>

Hope this helps

7 0

3 years ago

A 4.00g sample of helium has a volume of 24.4L at a temperature of 25.0oC and a pressure of 1.00 atm. The volume of the helium i

Masteriza [31]

Answer:

0.41 moles.

Explanation:

Given that:

Mass of helium = 4.00 g

Initial Volume = 24.4 L

initial Temperature = 25.0 °C =( 25 + 273) = 298 K

initial Pressure = 1.00 atm

The volume was reduced to :

i.e

final volume of the helium - 10.4 L

Change in ΔV = 24.4 - 10.4 = 10.0 L

Temperature and pressure remains constant.

The new quantity of gas can be calculated by using the ideal gas equation.

PV = nRT

n = $\frac{PV}{RT}$

n = $\frac{1.00*10.0}{0.082057*298}$

n = 0.4089 moles

n = 0.41 moles.

7 0

3 years ago

A calorimeter contains 500 g of water at 25°C. You place a hand warmer containing 100 g of liquid sodium acetate (NaAC) inside t

antoniya [11.8K]

The heat absorbed by the water is
Q = 500 (4.18) (32.2 - 25)
Q = 15048 J

The enthalpy of fusion of the sodium acetate is:
<span>ΔHf = Q / m
</span><span>ΔHf = 15048 / 100
</span>ΔHf = 150.48 J/g

3 0

3 years ago

Read 2 more answers

Consider the reaction between hydrogen gas and chlorine gas to form hydrogen chloride gas in a closed cylinder with a movable pi

d1i1m1o1n [39]

Answer:

In this case, the system doesn't be affected by the pressure change. This means that nothing will happen

Explanation:

We can answer this question applying the Le Chatelier's Principle. It says that changes on pressure, volume or temperature of an equilibrium reaction will change the reaction direction until it returns to the equilibrium condition again.

The results of these changes can define as:

Changes on pressure: the reaction will move depending the quantity of moles on each side of the reaction

Changes on temperature: The reaction will move depending on if it's endothermic or exothermic

Changes on volume: The reaction will move depending the limit reagent and the quantity of moles on each side of the reaction

In the exercise, they mention a change on pressure of the system at constant temperature (that means the temperature doesn't change). As Le Chatelier Principle's says, we must analyze what happens if the pressure increase or decrease. If pressure increase the reaction will move on the side that have less quantity of moles, otherwise, if the pressure decreases the reaction will move to the side that have more quantity of moles. In this case, we can see that both sides of the equation have the same number of moles (2 for the reactants and 2 for the products). So, in this case, we can conclude that, despite the change on pressure (increase or decrease), nothing will happen.

3 0

3 years ago

When 0.105 mol propane, C3H8 is burned in an excess of oxygen, how many moles of oxygen are consumed? The products are carbon di

torisob [31]

<u>Answer:</u> The amount of oxygen gas consumed is 0.525 moles

<u>Explanation:</u>

We are given:

Moles of propane burned = 0.150 moles

The chemical equation for the combustion of propane follows:

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

By Stoichiometry of the reaction:

1 mole of propane reacts with 5 moles of oxygen gas

So, 0.150 moles of propane will react with = $\frac{5}{1}\times 0.105=0.525mol$ of oxygen gas

Hence, the amount of oxygen gas consumed is 0.525 moles

6 0

3 years ago