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Andreyy89
3 years ago
14

Dose the weight of a human change

Chemistry
1 answer:
prisoha [69]3 years ago
4 0

Answer:

It changes as they grow

Or when they start to have an unhealthy diet

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2.A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a s
Aleks04 [339]

Answer:

In order to prepare 10 mL, 5 μM; <em> 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.</em>

In order to prepare 10 mL, 10 μM; <em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM; <em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM; <em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

Molarity x volume (initial)= Molarity x volume (final).

In order to prepare 10 mL, 5 μM from 25 μM solution,

Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?

25 x initial volume = 5 x 10

Initial volume = 50/25

                       = 2 mL

<em>2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

<em />

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

<em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

<em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

<em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

6 0
3 years ago
Write a word equation for the chemical reaction.<br><br> I’ll brainlist.
Alex

Answer:

potassium hydroxide + sulfuric acid → potassium sulfate + water.

Explanation:

Hope it helps.. if yes, plz mark me as brainliest

5 0
3 years ago
The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:
aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

                         H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium:  0.50M - x          x            x

The equilibrium constant (Ka₁) is:

K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M

Similarly, from the second dissociation of sulfurous acid we have:

                              HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium:  7.94x10⁻²M - x          x            x

The equilibrium constant (Ka₂) is:  

K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}  

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0
3 years ago
The lightest weight element that is not a gas
LUCKY_DIMON [66]

Answer:Lithium

Explanation:Lithium has 3 protons, and 4 neutrons making it the lightest element that isn't a gas.

5 0
3 years ago
Read 2 more answers
Which scientist determined the charge of the electron?.
Alexus [3.1K]

Answer:

J.J. Thomson

Explanation:

3 0
2 years ago
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