Malonate is an aggressive inhibitor of succinate dehydrogenase. If malonate is added to a mitochondrial education this is oxidizing pyruvate as a substrate, it is lower in attention<u> </u><u>Fumarate</u><u>.</u>
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Succinate dehydrogenase is also known as mitochondrial complicated II, and inhibition of succinate dehydrogenase by means of dimethyl malonate has been said to suppress the production of pro-inflammatory cytokines.
Fumaric acid is an organic compound with the system HO₂CCH=CHCO₂H. It has a fruit-like taste and has been used as a meal additive. . The salts and esters are referred to as fumarates. Fumarate also can consult with the C ₄H ₂O²⁻ ₄ ion.
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Answer:
The freshwater sources that are generally in continuous motion and follow a defined path are called streams and rivers.
If I were to improve the lab then I will make the following changes:
- The experiment aimed to observe and model the effects of rivers on erosion. So, I can make a virtual model of the river and can compare the velocity, gradients and volume of rivers.
- Comparison between the low and high factors listed can help in computing the effect of the powerful river on erosion.
- The high velocity. gradient and volume of the river will cause more erosion as it exerts more force.
- The low volume, gradient and velocity river will affect in a less manner on erosion.
Explanation:
thats all i know ( correct me if im wrong please)
The molar mass of lithium oxide is
29.88 g/mol
Q = mCΔT
Q is heat in joules, m is mass, C is specific heat, and delta T is change in temp
2099 J = (40.27g)(C)(148.5 - 24.8) = .421 J / gram K
Answer:
Kc = 8.05x10⁻³
Explanation:
This is the equilibrium:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
Initially 0.0733
React 0.0733α α/2 3/2α
Eq 0.0733 - 0.0733α α/2 0.103
We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.
Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.
3/2α = 0.103
α = 0.103 . 2/3 ⇒ 0.0686
So, concentration in equilibrium are
NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682
N₂ = 0.0686/2 = 0.0343
So this moles, are in a volume of 1L, so they are molar concentrations.
Let's make Kc expression:
Kc= [N₂] . [H₂]³ / [NH₃]²
Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³
