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Natasha_Volkova [10]
2 years ago
7

What color did you observing of marker?not color its a pure of in​

Chemistry
1 answer:
Gekata [30.6K]2 years ago
7 0
Blue and purple and red
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An enzyme is discovered that catalyzes the chemical reaction SAD ↔ HAPPY A team of motivated researchers sets out to study the e
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Answer:

The K_{m} of a substrate will be "10 μM".

Explanation:

The given values are:

E_{t} = 20 \ nM

[Substract] = 40 \ \mu M

K_{cat}=600 \ s^{-1}

Reaction velocity, Vo=9.6 \ \mu M s^{-1}

As we know,

⇒  Vo=\frac{K_{cat}[E_{t}][S]}{K_{m}+[S]}

On putting the estimated values, we get

⇒  9.6=\frac{600\times 20\times 10^{-3}\times 40}{K_{m}+40}

⇒  K_{m}+40=\frac{600\times 20\times 10^{-3}\times 40}{9.6}

⇒  K_{m}+40=50

On subtracting "40" from both sides, we get

⇒  K_{m}+40-40=50-40

⇒  K_{m}=10 \ \mu M

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3 years ago
Consider the following reaction. Zn(s) 2agno3(aq) --> 2ag(s) zn(no3)2(aq)when 16. 2 g of silver was produced, _____ mole(s) o
Crazy boy [7]

The required amount of silver nitrate to produce 16.2g of silver is 25.48 grams.

<h3>What is the relation between mass & moles?</h3>

Relation between the mass and moles of any substance will be represented as:

n = W/M, where

  • W = given mass
  • M = molar mass

Moles of silver = 16.2g / 107.8g/mol = 0.15mol

From the stoichiometry of the given reaction it is clear that, same moles of silver nitrate is required to produce same moles of silver. So 0.15 moles of silver nitrate is required.

Mass of silver nitrate = (0.15mol)(169.87g/mol) = 25.48g

Hence required mass of silver nitrate is 25.48g.

To know more about mass & moles, visit the below link:

brainly.com/question/19784089

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2 years ago
If you had excess chlorine, how many moles of of aluminum chloride could be produced from 19.0 g of aluminum?
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The chemical reaction would be written as follows:

2Al + 3Cl2 = 2AlCl3

We are given the amount of aluminum to be used in the reaction. This will be the starting point of the calculations. We do as follows:

19.0 g Al ( 1 mol / 29.98 g ) ( 2 mol AlCl3 / 2 mol Al ) = 0.63 mol AlCl3
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