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2 years ago

What is the hydronium concentration of a solution of ph 4.19M

Chemistry

1 answer:

V125BC [204]2 years ago

8 0

Answer:

.000064565

Explanation:

H3O+=10^-pH

- Hope that helps! Please let me know if you need further explanation.

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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

Pepsi [2]

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

⇒ $K_{m}+40-40=50-40$

⇒ $K_{m}=10 \ \mu M$

6 0

2 years ago

I really need help please

Greeley [361]

Answer:

I think it's B

hope this helps

5 0

3 years ago

What transition energy corresponds to an absorption line at 527 nm?

Ede4ka [16]

Answer:

E = 3.77×10⁻¹⁹ J

Explanation:

Given data:

Wavelength of absorption line = 527 nm (527×10⁻⁹m)

Energy of absorption line = ?

Solution:

Formula:

E = hc/λ

h = planck's constant = 6.63×10⁻³⁴ Js

c = speed of wave = 3×10⁸ m/s

by putting values,

E = 6.63×10⁻³⁴ Js × 3×10⁸ m/s / 527×10⁻⁹m

E = 19.89×10⁻²⁶ Jm /527×10⁻⁹m

E = 0.0377×10⁻¹⁷ J

E = 3.77×10⁻¹⁹ J

4 0

2 years ago

Read 2 more answers

Hi, my question is not answer, please

Oksanka [162]

Okay, pls give brainliest because I answered the fastest and how are you?

4 0

2 years ago

A student puts 0.020 mol of methyl methanoate into an empty and rigid 1.0 L vessel at 450 K. The pressure is measured to be 0.74

stellarik [79]

Explanation:

Starting moles of ethanol acid = 0.020 mol

At the equilibrium 50 % of the ethanol acid molecules reacted

∴ Moles of ethanol acid reacted = 0.020 mol * 50 %/100 %

= 0.010 mol

Moles of ethanol acid remain = 0.020 mol + 0.010 mol = 0.010 mol

Moles of the product $(CH3COOH)^{2}$ gas formed are calculated as

0.010 mol CH3COOH * 1 mol $(CH3COOH)^{2}$ / 2 mol CH3COOH

= 0.005 mol $(CH3COOH)^{2}$

Therefore at the equilibrium total moles of gas present in the vessel are 0.010 mol CH3COOH and 0.005 mol $(CH3COOH)^{2}$

That is total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol

Now Calculate the pressure :

0.020 mol gas has pressure of 0.74 atm therefore at the same condition what will be the pressure exerted by 0.015 mol gas

P1/n1 = P2/n2

P2 = P1*n2 / n1

= 0.74 atm * 0.015 mol / 0.020 mol

= 0.555 atm

4 0

3 years ago