Question

How Modern periodic classification outcomes the limitations of Mendeleev’s classification.

Answer 1

Answer:

Periodic Table

Limitations of the Mendeleev periodic table are listed below.

• Elements with large differences in properties were included in the same group. for example, hard metals like copper and silver were included along with soft metals like sodium and potassium.
• No proper position could be given to the element hydrogen. In the periodic table, the location of hydrogen is uncertain. It was put with alkali metals within 1A class, but certain hydrogen properties are close to those of halogens. So, it can also be put for halogens in the band.
• The increasing order of atomic mass was not strictly followed throughout. For example in cobalt and nickel & Tellurium and Indium.
• As isotopes are atoms of the same element having different atomic masses, they should have been given different positions while arranging them in the order of atomic mass. But this was not done.

Explanation:

Hope it may help you...

Answer 2

Answer:

l\begin{gathered} \red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  \:  \: TO \: \: FIND \:  \:  \:  \maltese }}}}} \\ \\ \huge \mathfrak{Derivative \: \: of } \\ \\ \bf \frac{x+cosx}{tanx} \end{gathered}✠   TOFIND   ✠✠   TOFIND   ✠Derivativeoftanxx+cosx

\color{Cyan}{\qquad \qquad \large \underline{ \pmb{{ \mathbb{ \maltese    \: REQUIRED \: \: INFO \:  \maltese }}}}}✠  REQUIREDINFO ✠✠  REQUIREDINFO ✠

\begin{gathered} \bigstar \: \: \: \mathfrak{ \underline{\huge Quotient \: \: Rule}} \\ \\ \bf \frac{d}{dx} \frac{u}{v} = \frac{v . \frac{du}{dx} - u \frac{dv}{dx} }{v {}^{2} } \\ \\ \bigstar \bf \: \: \: \frac{d}{dx} cosx = - sinx \\ \\ \bigstar \: \: \: \bf \frac{d}{dx} tanx = {sec}^{2} x\end{gathered}★QuotientRuledxdvu=v2v.dxdu−udxdv★dxdcosx=−sinx★dxdtanx=sec2x

\color{magenta}{ \large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  \:  \: SOLUTION \:  \:  \:  \maltese }}}}}✠   SOLUTION   ✠✠   SOLUTION   ✠

\begin{gathered} \bf \frac{d}{dx} ( \frac{x + cosx}{tanx} ) \\ \\ = \sf \frac{tanx. \frac{d}{dx} (x + cosx) - (x + cosx) \frac{d}{dx} tanx}{ {(tanx)}^{2} } \\ ( \bf \because Quotient \: Rule ) \\ \\ = \sf\frac{tanx(1 - sinx) - (x + cosx) {sec}^{2} x}{ {tan}^{2} x} \end{gathered}dxd(tanxx+cosx)=(tanx)2tanx.dxd(x+cosx)−(x+cosx)dxdtanx(∵QuotientRule)=tan2xtanx(1−sinx)−(x+cosx)sec2x

\begin{gathered} \sf = \frac{tanx \: - \: tanx.sinx \: - \: x {sec}^{2} x \: - \: cosx. {sec}^{2} x}{ {tan}^{2}x } \\ \\ =\sf \frac{tanx \: - \: tanx.sinx \: - \: x {sec}^{2} x \: - \: cosx. \dfrac{1}{ {cos}^{2}x } }{ {tan}^{2}x } \\ \\ = \bf\frac{tanx \: - \: tanx.sinx \: - \: x {sec}^{2} x \: - \: {sec}x}{ {tan}^{2}x } \end{gathered}=tan2xtanx−tanx.sinx−xsec2x−cosx.sec2x=tan2xtanx−tanx.sinx−xsec2x−cosx.cos2x1=tan2xtanx−tanx.sinx−xsec2x−secx

\large\red{ \mathfrak{ \text{W}hich \: \: is \: \text{T}he \: \: Required \: \: \text{ A}nswer }}WhichisTheRequired Answer