**Buffer **is a solution which **resist** the **change in pH** with addition of small amount of **acid** or **base**.

Hence when **small** amount of **acid** is **added **to **buffer** its **pH will not change,** as the buffer has the capacity to **resist **the change in pH on **small addition** of **acid or base** to it.

Buffer is mixture of **weak acid**/**base** and its **conjugate base**/**acid**.

Hence when **small amount of acid** is added to **HA/A-**. The **H+ ions** added reacts with the **conjugate base A- in buffer** forming the **weak acid** and thus keeping the **pH same**.

It's very simple... if we remember value of Universal Gas Constant R and Ideal Gas Law, so...

**Ideal Gas Law**

pV = nRT, where:

p - pressure (in kPa),

V - volume (in L),

n - number of moles (in mol),

R - universal cas constant (in kPa * L / mo l* K),

T - temperature (in K)

n = m/M, where:

n - number of moles,

m - mass (in grams),

M - molar mass of ingredient (in g/mol) - you find this at Periodic Table.

pV = nRT ---> pV = mRT/M ---> pVM = mRT ---> **pVM/RT = m**

p = 17615 kPa

T = 273.15 + 23 = 296.15 K

V = 43.8 L

R = 8.314 kPa * L / mol * K

M (for argon) = 39.948 g/mol

and

m = (17615 kPa * 48.3 L * 39.948 g/mol) / (296.15 K * 8.314 kPa * L / mol * K)

m = 13803.93 grams of Argon

**Answer:**

Hey mate....

**Explanation:**

This is ur answer....

**<em>1) All matter is made of atoms. Atoms are indivisible and indestructible. </em>****<em>3) Compounds are formed by a combination of two or more different kinds of atoms. 4) A chemical reaction is a rearrangement of atoms.</em>**

Hope it helps you,

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**The concentration of AB, in mol/dm³, after 6 minutes : 0.125**

<h3>

**Further explanation**</h3>

For first-order reaction :

The half-life of the reaction is 2 minutes⇒t1/2=2 minutes

The concentration of AB, in mol dm, after 6 minutes ⇒ t=6 minutes

The rate constant (k) :

The concentration after 6 minutes :

First look at the molar mass of Al, which is 26.98 grams. That means that there are 26.98 grams in one mole of Al. To find the moles in your given sample of 64.1 grams, just divide that value by the molar mass of 26.98 grams. That will give you a final answer of 2.38 moles in 64.1 grams of Al.

Hope this helps!