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2 years ago

What's the pH of a solution composed of 0.2 moles of NH3 and 0.15 moles of nh4cl

Chemistry

1 answer:

Norma-Jean [14]2 years ago

8 0

Answer:

$pH=4.63$

Explanation:

Hello!

In this case, since the ionization of ammonia, which is a weak base, is written as:

$NH_3+H_2O\rightleftharpoons NH_4^++OH^-$

We can see that the ammonium ion is the conjugate acid whereas the hydroxide ions the conjugate base; that is why we use the Henderson-Hasselbach equation to compute the pH, given the pKb of ammonia 4.75:

$pH=pKb+log(\frac{[conj\ acid]}{[base]} )$

In such a way, for the given moles of ammonia, base, and those of ammonium chloride, conjugate acid form, we obtain:

$pH=4.75+log(\frac{0.15}{0.2} )\\\\pH=4.63$

Best regards!

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Explain how carbon is conserved in the carbon cycle.

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Explanation:

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2 years ago

Write packing efficiency of fcc ,BCC, SCC and formula also . ?

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❖ Packing Efficiency ❖

➪ The percentage of total space occupied by particles is called packing efficiency.

$\\ \qquad{\rule{200pt}{3pt}}$

$\bigstar \boxed{ \sf \: \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}$

❖ Packing efficiency of simple cubic structure (SCC).❖

$\rm \: {Let \: } \bf{r } \: \rm{ be \: the \: radius \: of \: a \: sphere }\: \\ \rm and \: \bf{a} \: \rm {be \: the \: edge \: of \: unit \: cell.} \\ \\ \; \bigstar \boxed{ \sf{Volume_{(sphere)} = \dfrac{4}{3} \pi r^3}} \bigstar$

❒ Since, simple cubic unit cell contain 1 atom (sphere). So, the total volume of sphere will be :

$: : \implies \sf \: 1 \times \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi r^3$

Volume of unit cell = a³

Volume of unit cell = (2r)³

Volume of unit cell = 8r³

❒ Now, we know that,❒

$\bigstar \boxed{ \sf \: \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}$

➪ Substituting the known values in the formula, we get the following results:

$: : \implies \rm \: {\dfrac{\dfrac{4}{3}\pi r^3}{8r^3} \times 100} \\ \\ : : \implies \dfrac{\pi}{6} \times 100 \\\\\ : : \implies \bf {52.4 \%}$

❒ Hence, the packing efficiency of simple cubic structure is 52.4%.

$\\ \qquad{\rule{200pt}{3pt}}$

➪ Packing efficiency of cubic close packing (SCC)/face centred cubic structure (FCC).

❒ Since, simple cubic unit cell contain 2 atom (sphere). So, the total volume of sphere will be:

$: : \implies \rm 4 \times \dfrac{4}{3}\pi r^3 = \dfrac{16}{3}\pi r^3$

Volume of unit cell = a³

Volume of unit cell = (2√2r)³

Volume of unit cell = 16√2r³

❒ Now, we know that,❒

$\bigstar \boxed{ \sf \: \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}$

➪Substituting the known values in the formula, we get the following results:

$: : \implies \rm {\dfrac{\dfrac{16}{3}\pi r^3}{16\sqrt{2}r^3} \times 100 }\\\\ \implies \rm \dfrac{\pi}{3\sqrt{2}} \times 100 \\\\\ \implies\bf52.4 \%$

❖ Hence, the packing efficiency of face centred cubic structure is 74%.

$\\ \qquad{\rule{200pt}{3pt}}$

❖ Packing efficiency of body cubic structure (BCC).

❒ Since, simple cubic unit cell contain 2 atom (sphere). So, the total volume of sphere will be:

$: : \implies \rm \: 2 \times \dfrac{4}{3}\pi r^3 = \dfrac{8}{3}\pi r^3$

Volume of unit cell = a³

Volume of unit cell = (4r/√3)³

Volume of unit cell = 64r³/3√3

❒ Now, we know that,❒

$\bigstar \boxed{ \sf \: \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}$

➪Substituting the known values in the formula, we get the following results:

$: : \implies \rm {\dfrac{\dfrac{8}{3}\pi r^3}{\dfrac{64r^3}{3\sqrt{3}}} \times 100} \\\\ \implies \dfrac{3\pi}{8} \times 100 \\\ \\ \bf \implies \: 68 \%$

❒ Hence, the packing efficiency of body centred cubic structure is 68%.

$\\ \qquad{\rule{200pt}{3pt}}$

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Elements with atomic numbers of 104 and greater are known as super-heavy elements. None of these elements have been found in nat

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Here we have to justify the work of scientists to generate more unknown elements after atomic number 104 which are so unstable and cannot be obtained in nature.

It is fact that the stability of the super heavy element is hard to obtain, as when there are so much proton and neutron at the nucleus of such element makes it so unstable.

So the try to make more unknown element to enlarge the periodic table is not justified as per the stability and utility of such elements are concerned. But in fact it helps to get the new elements which may exist and may have some radioactivity which can be used.

But apparently the making of new material in laboratory to expand the periodic table is not justified.

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3 years ago

Read 2 more answers

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Answer:

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Explanation:

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3 years ago

What's the pH of a solution composed of 0.2 moles of NH3 and 0.15 moles of nh4cl​

What's the pH of a solution composed of 0.2 moles of NH3 and 0.15 moles of nh4cl