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Ainat [17]
2 years ago
10

After a product is released from its active site, a(n) ______ can be used again in another reaction because it is not consumed o

r altered in the reaction.
Chemistry
1 answer:
3241004551 [841]2 years ago
7 0

Answer:

Enzyme.

Explanation:

Enzymes act as a catalyst in all reaction but remain unchanged by the process.

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A compound containing phosphorus and oxygen has a molar mass of 219.9 g/mol and an empirical formula of p2o3. Determine its mole
Softa [21]

Answer:

P₄O₆

Explanation:

The molecular formula is a whole number multiple of the empirical formula. that is, if the mole wt is 219.9 gms/mole and the empirical formula weight is 110 gms/mole*, then the whole number multiple is 219.9/110 = 2 => Molecular formula => (P₂O₄)₂ => P₄O₆.

7 0
3 years ago
Which action best demonstrates the transformation of mechanical energy to heat energy?
Marysya12 [62]
I believe the answer is burning a candle
6 0
2 years ago
Read 2 more answers
What is the pH of a 0.00530 M solution of HCI?
MrMuchimi

Answer:

2.28

Explanation:

HCl(l) ===> H+ + cl-

HCl is a very strong acid. Almost all of it will decompose to the right. That means the concentration of H+ is 0.00530

pH = - log [H+]

pH = - log[0.00530]

pH = - - 2.2757

pH = 2.2757

Rounded this 2.28

5 0
3 years ago
A particular first-order reaction has a rate constant of 1.35 × 102 s-1 at 25.0°C. What is the magnitude of k at 95.0°C if Ea =
never [62]

Answer:

k ≈ 9,56x10³ s⁻¹

Explanation:

It is possible to solve this question using Arrhenius formula:

ln\frac{k2}{k1} = \frac{-Ea}{R} (\frac{1}{T2} -\frac{1}{T1} )

Where:

k1: 1,35x10² s⁻¹

T1: 25,0°C + 273,15 = 298,15K

Ea = 55,5 kJ/mol

R = 8,314472x10⁻³ kJ/molK

k2 : ???

T2: 95,0°C+ 273,15K = 368,15K

Solving:

ln\frac{k2}{k1} = 4,257

\frac{k2}{k1} = 70,593

{k2} = 9,53x10^3 s^{-1}

<em>k ≈ 9,56x10³ s⁻¹</em>

I hope it helps!

5 0
3 years ago
The gene for curled ears is dominant over the gene for straight ears (e). If you crossed a cat with curled ears (Ee) and a cat w
tatyana61 [14]

Answer:

50% is the answer I believe

8 0
3 years ago
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