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11111nata11111 [884]
3 years ago
13

Which is the energy source used to fuel photosynthesis?

Chemistry
2 answers:
ryzh [129]3 years ago
8 0
I think the answer is sunlight
8090 [49]3 years ago
3 0

Answer:

sunlight

Explanation:

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1; The change of matter from a solid state to a gas state
WINSTONCH [101]
<span>1 The change of matter from a solid state to a gas state is called sublimation.

</span><span>2. When a solute is completely dissolved in a solvent, a solution is formed. (A)

3. Base are used in </span><span>Used in ammonia, Alka-Seltzer, and dish soap
</span><span>
4. </span><span>TRUE . Simple devices lower the required input force.

5. FALSE. It's the opposite.</span>
6 0
3 years ago
2.1 g of a hydrocarbon fuel is burned in a calorimeter that contains 280 grams of water initially at 25.00◦C. After the combusti
olga2289 [7]

Answer:

\large \boxed{\text{933 J}}

Explanation:

There are three heat transfers involved.

heat from combustion of propane + heat gained by water + heat gained by calorimeter = 0

  q₁     +     q₂      +       q₃      = 0

m₁ΔH + m₂C₂ΔT + C_calΔT = 0

Data:

 m₁ =      2.1   g

 m₂ = 280     g

   Ti = 25.00 °C

   T_f = 26.55 °C

Ccal = 92.3 J·°C⁻¹

Calculations:

Let's calculate the heats separately.

1. q₁

q₁ = 2.1 g × ΔH = 2.1ΔH g

2. q₂

ΔT = T_f - Ti = 26.55 °C - 25.00 °C = 1.55 °C

q₂ = 280 g × 4.184 J·°C⁻¹ × 1.55 °C = 1816 J

3. q₃

q₃ = 92.3 J·°C⁻¹ × 1.55 °C = 143.1 J

4. ΔH

\begin{array}{rcl}\text{2.1$\Delta$H g +1816 J +143.1 J} & = & 0\\\text{2.1$\Delta$H g +1959 J} & = & 0\\\text{2.1$\Delta$H g}& = & \text{-1959 J}\\\Delta H & = & \dfrac{\text{-1959 J}}{\text{2.1 g}}\\\\& = & \textbf{-933 J/g}\\\end{array}\\\text{The combustion releases $\large \boxed{\textbf{933 J}}$ per gram of fuel burned.}

7 0
3 years ago
The key elements of geosphere are trees, rocks, and dirt.<br><br> True or False
Dimas [21]
Answer: This is true!
4 0
3 years ago
At a given temperature the vapor pressure of pure liquid benzene and toluene are 745 torr and 290 torr, respectively. A solution
IRISSAK [1]

Answer:

Vapour pressure of benzene over the solution is 253 torr

Explanation:

According to Raoult's law for a mixture of two liquid component A and B-

vapour pressure of a component (A) in solution = x_{A}\times P_{A}^{0}

vapour pressure of a component (B) in solution = x_{B}\times P_{B}^{0}

Where x_{A},x_{B} are mole fraction of component A and B in solution respectively

P_{A}^{0},P_{B}^{0} are vapour pressure of pure A and pure B respectively

Here mole fraction of benzene in solution is 0.340 and vapour pressure of pure benzene is 745 torr

So, vapour pressure of benzene in solution = 0.340\times 745 torr

                                                                         = 253 torr

7 0
3 years ago
Need help with this can some please help
vekshin1
I believe you would just put a 2 in front of NH3 and keep the other ones as 1
7 0
3 years ago
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