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matrenka [14]
3 years ago
8

How many ligaments in e kne

Chemistry
1 answer:
Makovka662 [10]3 years ago
7 0
4.
Medial collateral ligament, which runs along inside of the knee.
Lateral collateral ligament, which runs along the outside of the knee.
Ligament, which runs in the middle part of the knee.
Posterior cruciate ligament, which works together with the Anterior cruciate ligament.
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Review the PCR (Calvin-Benson-Basham) cycle and the photochemical (light) reactions. If photosynthesizing green algae are provid
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Explanation:

THE PHOTOCHEMICAL(LIGHT) REACTIONS :This is a phase of photosynthesis where sunlight is used as a source of energy to manufacture two chemical compounds which are "Reduced nicotinamide adenine dinucleotide phosphate"-NADPH and "Adenosine triphosphate"-ATP.This phase of photosynthesis involves 4 steps or reactions which are :

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STEP 3:Formation of NADPH -The hydrogen ions (H+) produced reacts with an NADP ( an electron carrier in the chlorophyll) to form NADPH.

STEP 4: FORMATION OF ATP - The high energy generated from the electron transfer process or chain is used to add a phosphate group to ADP (Adenosine dphosphate) to form ATP.

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3 years ago
How does Lewis Dot Structures can be used to show reactions between different chemical elements​
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I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol
galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = \Delta H_f_{(I_2)}=62.438 kJ/mol

Enthalpy of formation of chlorine gas = \Delta H_f_{(Cl_2)}=0 kJ/mol

Enthalpy of formation of ICl gas = \Delta H_f_{(ICl)}=17.78 kJ/mol

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]

=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol

Enthaply change when 1.62 moles of iodine gas recast:

\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ

Entropy of the surrounding = \Delta S^o_{surr}=\frac{\Delta H}{T}

=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

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The s orbital can hold
malfutka [58]

Answer:

B. Two electrons

Explanation:

Since the S orbital is closest to the nucleus and it only has one orbital, it can only hold 2 electrons

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