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fenix001 [56]
3 years ago
9

In which way are photosynthesis and cellular respiration different?

Chemistry
2 answers:
tangare [24]3 years ago
7 0
The correct answer is D
Lyrx [107]3 years ago
3 0

Answer:

D. Photosynthesis uses carbon dioxide while cellular produces carbon dioxide

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What does nitrogen do to the soil
Tresset [83]

Answer:The chemical element will always get converted during the cycle as it enters different ecosystems. Nitrogen plays an important role as a nutrient in soils. It is needed for photosynthesis in plants. Nitrogen helps decomposers such as bacteria, worms

Explanation:

3 0
3 years ago
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1 C8H10(l) +21/2O2(g) → 8CO2(g) + 5H2O(g), Hcomb= ? Hf for C8H10(l) = +49.0kJ/mol C8H10(l) Use the balanced combustion reaction
nikklg [1K]

Answer:

H_{comb}=-4406kJ/mol

Explanation:

Hello,

In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):

H_{comb}=8*\Delta _fH_{CO_2}+5\Delta _fH_{H_2O}-\Delta _fH_{C_8H_{10}}

Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:

H_{comb}=8*(-393.5kJ/mol)+5(-241.8kJ/mol)-49.0kJ/mol\\\\H_{comb}=-4406kJ/mol

Best regards.

4 0
3 years ago
What is the energy of a 933 nm wave?
vovikov84 [41]

E=hc/l

E=

<span><span>E=<span>(6.626 x 10-34 J s)(3.0 x 108m/s )</span><span>=2.88 x 10-19J</span></span><span>6.90 x 10-7m</span></span>
8 0
2 years ago
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Transferring or sharing electrons between atoms forms
Brums [2.3K]
Transferring or sharing electrons between atoms forms a covalent bond.<span> Covalent bonding is when atoms share electrons. It is a chemical bond that involves the sharing of electron pairs. These pairs are called bonding pairs. Examples of compounds that has covalent bonds are CO2, organic compounds, lipids and proteins.</span>

6 0
3 years ago
Assuming that the experiments performed in the absence of inhibitors were conducted by adding 5 μl of a 2 mg/ml enzyme stock sol
prohojiy [21]

Hey there!:

From the given data ;

Reaction  volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )

Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:

[E]t in molar  = g/L * mol/g

[E]t  = 0.01 g/L * 1 / 45,000

[E]t = 2.22*10⁻⁷

Vmax = 0.758 umole/min/ per mL

= 758 mmole/L/min

=758000 mole/L/min => 758000 M

Therefore :

Kcat = Vmax/ [E]t

Kcat = 758000 / 2.2*10⁻⁷ M

Kcat = 3.41441 *10¹² / min

Kcat = 3.41441*10¹² / 60 per sec

Kcat = 5.7*10¹⁰ s⁻¹

Hence   kcat of   xyzase is  5.7*10¹⁰ s⁻¹


Hope that helps!



4 0
3 years ago
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