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jasenka [17]
3 years ago
5

How do you write 561983 in scientific notation?

Chemistry
1 answer:
Artist 52 [7]3 years ago
7 0

Answer:

5.61983 × 10^5

Explanation:

Move the decimal forward 5 spaces, each time doing so you add 10^(# of spaces moved, in this case 5)

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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
Describe 3 differences between ice, liquid water and steam.
Evgen [1.6K]

Steam is a gas. ... The difference between a liquid and the other states of matter is that liquid molicules are more spread apart than solid molicules but less spread apart than gas molicues. Water is the base for both ice and steam.on:

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3 years ago
PLEASE HELP ME!
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The pH scale is used to measure the degree of acidity or alkalinity of a solution. The scale runs from 0 (very acidic solutions can have a negative pH) to 14 (very alkaline solutions can have a pH higher than this), while a neutral liquid such as pure water has a pH of 7. The pH is linked to the concentration of hydrogen ions (H +) in the solution. Diluting an acid or alkali affects the concentration of H +<span> ions in a solution and therefore affects the pH. In this activity, we will investigate how diluting an acid or alkali affects the pH.
Hope this helps:D
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8 0
3 years ago
Question 3
meriva

Answer:

We are dependent on plants and plants need CO2 from enviromen

Explanation:

8 0
3 years ago
Read 2 more answers
Determine the value of the equilibrium constant, Kgoal, for the reaction C(s)+12O2(g)+H2(g)⇌12CH3OH(g)+12CO(g), Kgoal=? by makin
Licemer1 [7]

Answer:

1.71x10²⁷

Explanation:

If we sum 1/2 of (3) + 1/2 of (1):

1/2 (3.) C(s) + 1/2O₂(g) ⇌ CO(g), K₃ = √2.10×10⁴⁷  = 4.58x10²³

1/2 (1)   1/2CO₂(g) + 3/2H₂(g) ⇌ 1/2CH₃OH(g) + 1/2H₂O(g), K₁ = √1.40×10² = 11.8

C(s) + 1/2O₂(g) +<u> 1/2CO₂(g) </u>+<u> 3/2H₂(g</u>) ⇌ 1/2CH₃OH(g) + <u>1/2H₂O(g)</u> + <u>CO(g)</u>

K' = 4.58x10²³ * 11.8 = 5.42x10²⁴

+1/2 (2):

<u>1/2 CO(g)</u> +<u> 1/2H₂O(g)</u> ⇌<u> 1/2CO₂(g)</u> + <u>1/2H₂</u> (g), K = √1.00×10⁵ = 316.2

C(s) + 1/2O₂(g) + H₂(g) ⇌ 1/2 CHO₃H(g) + 1/2CO(g)

K'' = 5.42x10²⁴* 316.2 =

<h3>1.71x10²⁷</h3>

5 0
4 years ago
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