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2 years ago

Can anyone teach me how to do acid and base? (chemistry)

Chemistry

2 answers:

NARA [144]2 years ago

3 0

Acids have a PH level of 6 or lower
Bases have a PH of 8 or higher
PH 7 is neutral

elena55 [62]2 years ago

3 0

For the topic of conjugate acids and bases you know that the if you have an acid (sometimes labeled as HA), its conjugate base is HA minus one H⁺ ion (sometimes called a proton). The conjugate base of HCl is Cl₋, the conjugate base of H₂SO₄ is HSO₄⁻, the conjugate base of CH₃COOH is CH₃COO⁻, and so on. To find the conjugate acid of a base, you have to reverse that process. The conjugate acid of HCO₃⁻ is H₂CO₃, the conjugate acid of NH₃ is NH₄⁺, the conjugate acid of PO₄³⁻ is HPO₄²⁻, and so on.

examples:
HCl+H₂O⇒H₃O⁺+Cl⁻
acid=HCl, base=H₂O, conjugate base=Cl⁻, and conjugate acid=H₃O⁺

NH₃+H₂O⇒NH₄⁺+OH⁻
acid=H₂O, base=NH₃, conjugate base=OH⁻, and conjugate acid=NH₄⁺

I hope this helps. Let me know if anything is unclear or you want me to explain anything further.

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Would someone mind helping me? I really need this answer but I'm so confused. I would appreciate any help :) and if you get the

Svetllana [295]

Answer:

Explanation:

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2 years ago

Read 2 more answers

Identify the species oxidized. 2 al3+(aq) + 2 fe(s) → 2 al(s) + 3 fe2+(aq)

Sliva [168]

To know the answer, compare the oxidation number of the element in the reactant and the product side. The oxidation number of Al was originally +3, then became 0 after the reaction. On the other hand, Fe was originally 0, then became +2 after the reaction. When the element is oxidized, it oxidation number increases. <em>Thus, the element oxidized is Fe.</em>

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3 years ago

An alloy with an average grain diameter of 35 μm has a yield strength of 163 Mpa, and when it undergoes strain hardening, the gr

Lera25 [3.4K]

Answer:

$\sigma_y\ =210.2\ MPa$

Explanation:

Given that

d= 35 μm ,yield strength = 163 MPa

d= 17 μm ,yield strength = 192 MPa

As we know that relationship between diameter and yield strength

$\sigma_y=\sigma_o+\dfrac{K}{\sqrt d}$

$\sigma_y\ =Yield\ strength$

d = diameter

K =Constant

$\sigma_o\ =material\ constant$

So now by putting the values

d= 35 μm ,yield strength = 163 MPa

$163=\sigma_o+\dfrac{K}{\sqrt 35}$ ------------1

d= 17 μm ,yield strength = 192 MPa

$192=\sigma_o+\dfrac{K}{\sqrt 17}$ ------------2

From equation 1 and 2

$192-163=\dfrac{K}{\sqrt 17}-\dfrac{K}{\sqrt 35}$

K=394.53

By putting the values of K in equation 1

$163=\sigma_o+\dfrac{394.53}{\sqrt 35}$

$\sigma_o\ =96.31\ MPa$

$\sigma_y=96.31+\dfrac{394.53}{\sqrt d}$

Now when d= 12 μm

$\sigma_y=96.31+\dfrac{394.53}{\sqrt 12}$

$\sigma_y\ =210.2\ MPa$

4 0

3 years ago

A ball is rolled along and off a table. Which of the following acts on the ball after it leaves the

alexira [117]

Both the force of the earth’s gravity on the ball and the force the ball got from being rolled off

6 0

2 years ago

A critical reaction in the production of energy in biological systems is the hydrolysis of adenosine triphosphate (ATP) to adeno

Archy [21]

Answer:

ΔG° of reaction = -47.3 x $10^{3}$ J/mol

Explanation:

As we can see, we have been a particular reaction and Energy values as well.

ΔG° of reaction = -30.5 kJ/mol

Temperature = 37°C.

And we have to calculat the ΔG° of reaction in the biological cell which contains ATP, ADP and HPO4-2:

The first step is to calculate the equilibrium constant for the reaction:

Equilibrium Constant K = $\frac{[HPO4-2] x [ADP]}{ATP}$

And we have values given for these quantities in the biological cell:

[HP04-2] = 2.1 x $10^{-3}$ M

[ATP] = 1.2 x $10^{-2}$ M

[ADP] = 8.4 x $10^{-3}$ M

Let's plug in these values in the above equation for equilibrium constant:

K = $\frac{[2.1x10^{-3}] x [8.4x10^{-3}] }{[1.2 x 10^{-2}] }$

K = 1.47 x $10^{-3}$ M

Now, we have to calculate the ΔG° of reaction for the biological cell:

But first we have to convert the temperature in Kelvin scale.

Temp = 37°C

Temp = 37 + 273

Temp = 310 K

ΔG° of reaction = (-30.5 $10^{3}$ ) + (8.314)x (310K)xln(0.00147)

Where 8.314 = value of Gas Constant

ΔG° of reaction = (-30.5 x $10^{3}$ ) + (-16810.68)

ΔG° of reaction = -47.3 x $10^{3}$ J/mol

5 0

2 years ago