Question

What is the mass in gram of 0.5 mole of nitric acid??​

Answer 1

Answer:

$\large\pink{\boxed{ \green{\rm 31.5 \: grams}}}$

Explanation:

Have Given :-

$\rm nitric \: acid \: (HNO_3) = 0.5 \: mole$

$\rm Molecular \: weight \: of \: HNO_3 : -$

$\implies1 + 14 + 16 \times 3$

$\implies15 + 48$

$\rm \implies63\:grams$

$\rm Number \: of \: moles (n)= \frac{weight}{molecular \: weight}$

$\rm Weight=n×Molecular  \: weight$

$= 0.5 \times 63$

$\rm = 31.5 \: grams$

Therefore mass of 0.5 mole of nitric acid is 31.5 grams

Answer 2

Answer:

Answer:

sinα - αcosα

Step-by-step explanation:

Given that,

\lim_{x\to\alpha}\;\;|\frac{x\sin\alpha-\alpha\sin x}{x-\alpha}|lim

x→α

∣

x−α

xsinα−αsinx

∣

Let x - α = h

Thus when x ⇒ α , h ⇒ 0

\lim_{h\to0}\;\;|\frac{(h+\alpha)\sin\alpha-\alpha\sin(h+\alpha)}{h}|lim

h→0

∣

h

(h+α)sinα−αsin(h+α)

∣

Now we will evaluate left hand limit and right and limit separately

For LHL:-

\begin{gathered}\begin{gathered}\lim_{h\to0}\;\;\frac{\alpha\sin\alpha[1-\cos(-h)]}{-h}+\sin\alpha-\frac{\alpha\sin(-h)\cos\alpha}{-h}\\\;\\=\lim_{h\to0}\;\;\frac{\alpha\sin\alpha(1-\cosh)}{-h}+\sin\alpha-\frac{\alpha\sin h\cos\alpha}{h}\\\;\\=\alpha\sin\alpha[\lim_{h\to0}\frac{(1-\cosh)}{-h}]+\sin\alpha-\alpha\cos \alpha[\lim_{h\to0}\frac{\sin h}{h}]\\\;\\=0+\sin\alpha-\alpha\cos\alpha\\\;\\=\sin\alpha-\alpha\cos\alpha\end{gathered} \end{gathered}

h→0

lim

−h

αsinα[1−cos(−h)]

+sinα−

−h

αsin(−h)cosα

=

h→0

lim

−h

αsinα(1−cosh)

+sinα−

h

αsinhcosα

=αsinα[

h→0

lim

−h

(1−cosh)

]+sinα−αcosα[

h→0

lim

h

sinh

]

=0+sinα−αcosα

=sinα−αcosα

For RHL:-

\begin{gathered}\begin{gathered}\lim_{h\to0}\;\;\frac{\alpha\sin\alpha[1-\cosh]}{h}+\sin\alpha-\frac{\alpha\sinh\cos\alpha}{h}\\\;\\=\lim_{h\to0}\;\;\frac{\alpha\sin\alpha(1-\cosh)}{h}+\sin\alpha-\frac{\alpha\sin h\cos\alpha}{h}\\\;\\=\alpha\sin\alpha[\lim_{h\to0}\frac{(1-\cosh)}{h}]+\sin\alpha-\alpha\cos \alpha[\lim_{h\to0}\frac{\sin h}{h}]\\\;\\=0+\sin\alpha-\alpha\cos\alpha\\\;\\=\sin\alpha-\alpha\cos\alpha\end{gathered} \end{gathered}

h→0

lim

h

αsinα[1−cosh]

+sinα−

h

αsinhcosα

=

h→0

lim

h

αsinα(1−cosh)

+sinα−

h

αsinhcosα

=αsinα[

h→0

lim

h

(1−cosh)

]+sinα−αcosα[

h→0

lim

h

sinh

]

=0+sinα−αcosα

=sinα−αcosα

Thus limit is (sinα - αcosα).