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mestny [16]
3 years ago
9

Classify each phrase according to whether it applies to photophosphorylation, oxidative phosphorylation, or both

Chemistry
1 answer:
chubhunter [2.5K]3 years ago
6 0

Answer:

1. Both

2. Phosphorylation

3. Both

4. Phosphorylation

5. Oxidative.

6. Both

Explanation:

Phosphorylation only occurs in chloroplast and it involves larger electrical component. Both Phosphorylation and oxidative occurs in mitochondria and it involves proton gradient. They occur in plants to produce ATP. Oxidative involves in smaller electrical component.

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Draw the aromatic compound toluene (methylbenzene). show all hydrogen atoms, including those on the ring.
Otrada [13]

Aromatic compound has continuous cyclic structure with( 4n+2)π electrons (Huckels rule), where n = 0,1,2…

Here number of pi electron are 6, where 4 from two double bond and 2 from nitrogen non-bonding electrons, hence it has total 6 pi electrons, therefore

6= ( 4n+2)π

4 = 4n  

n =1

 

Hence it is an aromatic compound  

 


7 0
3 years ago
hat is the product when magnesium reacts with nitrogen? Mg(s) + N2(g) → Mg2N3(s) Mg3N(s) Mg3N2(s) MgN3(s)
Ipatiy [6.2K]
3Mg + N₂ = Mg₃N₂

Mg₃N₂
______________________
5 0
3 years ago
Read 2 more answers
In a sample of solid ba(no3)2 the ratio of barium ions to nitrate ions is
Nata [24]
I believe the ratio is 1:2
7 0
3 years ago
A 19-g piece of metal absorbs 186.75 joules of heat energy, and its temperature changes from 35°C to 175°C. Calculate the specif
slavikrds [6]

Answer:

0.0702J/g°C the specific heat capacity of the metal.

Explanation:m

Q=m\times c\times \Delta T=m\times c\times (T_{2}-T_{1})

where,

Q = heat absorbed by metal = 186.75 J

m_1 = Mass of metal= 19 g

T_1 = Initial  temperature of metal = 35^oC

T_2 =Final  temperature of metal = 175^oC

c = specific heat of metal= ?

186.75 J=19 g\times c\times (175^oC-35^oC)

c=\frac{186.75 J}{19 g\times (175^oC-35^oC)}

c=0.0702J/g^oC

0.0702J/g°C the specific heat capacity of the metal.

6 0
3 years ago
A solution is created by dissolving 13.5 grams of ammonium chloride in enough water to make 235 ml of solution. how many moles o
erica [24]
<span>Moles = 0.252 Molarity = 1.07 This question is badly worded. You're asking for moles and I suspect you really want molarity. The number of moles of ammonium chloride you have in the solution will remain constant regardless of the volume of the solution. However, the molarity of the solution will differ depending upon how concentrated it is. So I'll give you both the number of moles of ammonium chloride you have, and the molarity of the resulting solution. Please talk to your teacher if you're confused by the difference between moles and molarity. The formula for ammonium chloride is NH4Cl. So let's calculate it's molar mass. Start by looking up the associated atomic weights. Atomic weight nitrogen = 14.0067 Atomic weight hydrogen = 1.00794 Atomic weight chlorine = 35.453 Molar mass NH4Cl = 14.0067 + 4 * 1.00794 + 35.453 = 53.49146 g/mol Moles NH4Cl = 13.5 g / 53.49146 g/mol = 0.252376735 mol Molarity is defined as moles per liter, so let's divide the number of moles we have by the volume in liters. So: 0.252376735 mol / 0.235 l = 1.073943551 M Rounding to 3 significant figures gives: 0.252 moles, 1.07 molarity.</span>
7 0
3 years ago
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