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3 years ago

What is the purpose of using extract from chloroplasts?

Chemistry

2 answers:

Charra [1.4K]3 years ago

3 0

Answer:

When a light is shone on the extract, pigment molecules absorb energy. Because the pigments have been isolated from the thylakoid membranes of the chloroplasts, the energy cannot be used for photosynthesis. Instead, the energy is released as heat and light in a process called fluorescence.

vlabodo [156]3 years ago

3 0

Answer:

As for your final task for this lesson, you will need to prepare and deliver a speech on a familiar issue. You are free to decide on what issue to discuss, but your speech should only range two (2) to three (3) minutes only. You will also have to employ the different techniques in public speaking cited in this lesson.

Explanation:

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A solution is 40.00% by volume benzene (C6H6) in carbon tetrachloride at 20°C. The vapor pressure of pure benzene at this temper

finlep [7]

Answer:

The total vapor pressure is 84.29 mmHg

Explanation:

Step 1: Data given

Solution = 40.00 (v/v) % benzene in CCl4

Temperature = 20.00 °C

The vapor pressure of pure benzene at 20.00 °C = 74.61 mmHg

Density of benzene is 0.87865 g/cm3

The vapor pressure of pure carbon tetrachloride is 91.32 mmHg

We suppose the total volume = 100 mL

Step 2: Calculate volume benzene and CCl4

40 % benzene = 40 mL

60 % mL CCl4 = 60 mL

Step 3: Calculate mass benzene

Mass = density * volume

Mass of benzene = 40.00 mL * 0.87865 g/mL = 35.146 g

Step 4: Calculate moles of benzene

Moles = mass / molar mass

Number of moles of benzene = 35.146 grams / 78 g/mol = 0.45059 mol

Step 5: Calculate mass of CCl4

Mass of CCl4 = 60 mL * 1.5940 g/mL = 95.64 g

Step 6: Calculate moles CCl4

Number of moles of CCl4 = 95.64 grams / 154g/mol = 0.62104 mol

Step 7: Calculate total number of moles

Total number of moles = moles benzene + moles CCl4

0.45059 moles + 0.62104 moles = 1.07163 mol

Step 8: Calculate mole fraction benzene and CCl4

Mole fraction = moles benzene / total moles

Mole fraction of benzene = 0.45059 / 1.07163 = 0.4205

Mole fraction of CCl4 = 0.62104 / 1.07163 = 0.5795

Step 9: Calculate partial pressure

Partial pressure of benzene = 0.4205 * 74.61 = 31.37 mmHg

Partial pressure of CCl4 = 0.5795 * 91.32 = 52.92 mmHg

Total vapor pressure = 31.37 + 52.92 = 84.29 mmHg

The total vapor pressure is 84.29 mmHg

7 0

3 years ago

Which option correctly describes the relative charges and masses of the subatomic particles?

Alina [70]

Answer:

Explanation:

D Is The Answer Fella, My Head Hurt Really Bad

3 0

3 years ago

Which of the following is a chemical property?

ad-work [718]

The first one is D hope it helps!

6 0

3 years ago

Read 2 more answers

Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2O(g) (Delta

shusha [124]

Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

Explanation:

The balanced chemical reaction is,

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]$

Putting the values we get :

$\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]$

$\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]$

$\Delta H=-5314.8kJ$

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat = $\frac{5314.8}{2}\times 1=2657.4kJ$

Thus enthalpy of combustion per mole of butane is -2657.4 kJ

3 0

3 years ago

If I need 2.2 moles of CO2 , and I have excess Fe2O3 , how many moles of C do I need?

olchik [2.2K]

Answer:

0.733 mol.

Explanation:

From the balanced equation:

2Fe₂O₃ + C → Fe + 3CO₂,

It is clear that 1.0 moles of Fe₂O₃ react with 1.0 mole of C to produce 1.0 mole of Fe and 3.0 moles of CO₂.

Since Fe₂O₃ is in excess, C will be the limiting reactant.

Using cross multiplication:

1.0 mole of C produces → 3.0 moles of CO₂, from the stichiometry.

??? mole of C produces → 2.2 moles of CO₂.

∴ The no. of moles of C needed to produce 2.2 moles of CO₂ = (1.0 mole of C) (2.2 mole of CO₂) / (3.0 mole of CO₂) = 0.733 mol.

6 0

3 years ago