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Hunter-Best [27]
2 years ago
7

_______ is the process in which light energy is transformed into food energy.

Chemistry
1 answer:
ASHA 777 [7]2 years ago
8 0

Answer:

Photosynthesis

Explanation:

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A container of what volume would be required to store 60.0 moles of a gas at a pressure of 20.0 atm and a temperature of 50.0 Ce
belka [17]

Answer:

The answer to your question is:  79.6 L

Explanation:

Data:

n = 60 moles

Pressure = P = 20.0 atm

Temperature = T = 50°C

Volume = V = ?

Convert temperature to Kelvin

      °K = 273 + 50

      °K = 323

Formula  (ideal gas law)

               PV = nRT        solve for V

              V = nRT / P

             V = (60)(0.08206)(323) / 20        Substitution

            V = 1590.32 /20                    

             V = 79.6 L

               

7 0
4 years ago
What will be the theoretical yield of tungsten(is) ,W, if 45.0 g of WO3 combines as completely as possible with 1.50 g of H2
Anarel [89]

Answer:

35.6 g of W, is the theoretical yield

Explanation:

This is the reaction

WO₃  +  3H₂  →   3H₂O  +  W

Let's determine the limiting reactant:

Mass / molar mass = moles

45 g / 231.84 g/mol = 0.194 moles

1.50 g / 2 g/mol = 0.75 moles

Ratio is 1:3. 1 mol of tungsten(VI) oxide needs 3 moles of hydrogen to react.

Let's make rules of three:

1 mol of tungsten(VI) oxide needs 3 moles of H₂

Then 0.194 moles of tungsten(VI) oxide would need (0.194  .3) /1 = 0.582 moles (I have 0.75 moles of H₂, so the H₂ is my excess.. Then, the limiting is the tungsten(VI) oxide)

3 moles of H₂ need 1 mol of WO₃ to react

0.75 moles of H₂ would need (0.75 . 1)/3 = 0.25 moles

It's ok. I do not have enough WO₃.

Finally, the ratio is 1:1 (WO₃ - W), so 0.194 moles of WO₃ will produce the same amount of W.

Let's convert the moles to mass (molar mass  . mol)

0.194 mol . 183.84 g/mol = 35.6 g

3 0
3 years ago
Why does radius increase as you go down a group (hint: think Bohr-Rutherford diagrams)? Why would an increase in radius make it
harina [27]

Answer:

i rlly dont know i just want pionts pls follow me btw

4 0
3 years ago
Ordinary water boils at 100°C. Can it be made to boil at 95°C or 105°C.​
slava [35]

It can be done. Normally the boiling point of water is 100°C. It will boil at temperature greater than 100°C more quickly. Water can be boiled at 95°C but for that the atmospheric pressure of the water should be decreased which will decrease the boiling point of water.

<h3>Concept :</h3>

To boil water at 95°C, decrease the atmospheric pressure.

At 105°C, the water will be boiling quickly than normal at 100°C.

4 0
2 years ago
Read 2 more answers
in heating a kettle of water on an electric stove, 3.34×10^3 J of thermal energy was provided by the element of the stove. yet,
insens350 [35]

Answer:

The percentage efficiency of the electrical element is approximately 82.186%

Explanation:

The given parameters are;

The thermal energy provided by the stove element, H_{supplied} = 3.34 × 10³ J

The amount thermal energy gained by the kettle, H_{absorbed}  = 5.95 × 10² J

The percentage efficiency of the electrical element in heating the kettle of water, η%, is given as follows;

\eta \% = \dfrac{H_{supplied} - H_{absorbed} }{H_{supplied}}  \times 100

Therefore, we get;

\eta \% = \dfrac{3.34 \times 10^3 - 5.95 \times 10^2}{3.34 \times 10^3}  \times 100 = \dfrac{549}{668} \times 100 \approx 82.186 \%

The percentage efficiency of the electrical element, η% ≈ 82.186%.

4 0
3 years ago
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