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emmasim [6.3K]
4 years ago
12

Which substances pass through a leaf's stomata?

Chemistry
2 answers:
Vilka [71]4 years ago
7 0
Carbon dioxide pass through a leaf stomata.
Orlov [11]4 years ago
6 0

Oxygen can pass through the leafs stomata

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PLease answer the thing and explain why is that...
iragen [17]

Answer: i think its d

Explanation:

4 0
3 years ago
Determine the equilibrium constant, Kp, for the following reaction, by using the two reference equations below: 2 NO(g) + O2(g)
klio [65]

Answer:

Kp=3.07x10^6

Explanation:

Hello,

In this case, by knowing the given reference reactions, one could rearrange them as follows:

2 NO(g) \leftrightarrow N_2(g) + O_2(g); Kp_2 = \frac{1}{2.3 x 10^{-19}}=4.35x10^{18}

N_2(g) + 2O_2(g) \leftrightarrow 2NO_2(g);Kp_3=(8.4x10^{-7})^2=7.056x10^{-13}

Subsequently, to obtain the main reaction, we add the aforementioned reference rearranged reactions as shown below (just as reference):

2NO(g)+N_2(g)+2O_2\leftrightarrow 2NO_2(g)+N_2+O_2

Consequently, the equilibrium constant is computed as:

Kp=\frac{[N_2][O_2]}{[NO]^2} * \frac{[NO_2]^2}{[N_2][O_2]^2} =Kp_2*Kp_3=4.35x10^{18}*7.056x10^{-13}=3.07x10^6

Best regards.

8 0
4 years ago
What is the therm energy of an object?
Elan Coil [88]

Answer:

B

Explanation:

B. the total kinetic and potential energies of its particles

3 0
3 years ago
A 0.5438 g of a C.H.O. compound was combusted in air to make 1.039 g of CO2 and 0.6369 g H20. What is the empirical formula? Bal
goblinko [34]

Answer:

C₃H₅O₂

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

Explanation:

The reaction can be expressed as:

CₓHₓOₓ + nO₂ → CO₂ + H₂O

Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so <u>the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂</u>:

1.039gCO_{2}*\frac{1molCO_{2}}{44gCO_{2}} *\frac{1molC}{1molCO_{2}} *\frac{12gC}{1molC} =0.2834gC

All of the hydrogens atoms in the compound ended up becoming H₂O, so <u>the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O</u>:

0.6369g*\frac{1molH_{2}O}{18gH_{2}O} *\frac{1molH}{1molH_{2}O} *\frac{1gH}{1molH} =0.0354gH

Because the compound is composed only by C, H and O, <u>the mass of O in the compound can be calculated by substraction</u>:

0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O

In order to determine the empirical formula, we calculate the moles of each component:

  • mol C = 0.2834 g C ÷ 12 g/mol = 0.0236 mol C
  • mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H
  • mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O

Then we divide those values by the lowest one:

0.0236 mol C ÷ 0.0141 = 1.67

0.0354 mol H ÷ 0.0141 = 2.51

0.0141 mol O ÷ 0.0141 = 1

If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.

  • The reaction is:

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

8 0
3 years ago
How do potential energy diagrams represent energy transfer in<br> chemical reactions?
Digiron [165]

Answer:

Hope this helps

Explanation:

Potential energy diagrams represent the energy transfer in chemical reactions  in a diagram called a potential energy graph and/or a reaction progress curve. A potential energy diagram shows the adjustment in potential energy of a system as reactants are changed.

5 0
3 years ago
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