Solve the following equations
Question: If $A=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$, then $A A^{T}=$ _________ Solution: The given matrix is $A=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$. $\therefore A A^{T}$ $=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]^{T}$ $=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]\left[\begin{array}{lll}1 2 3\end{array}\right]$ $=\left[\begin{array}{lll}1 \times 1 1 \times 2 1 \times 3 \\ 2 \times 1 2 \times 2 ...
Read More →Which has more number of atoms?
Question: Which has more number of atoms? $100 \mathrm{~g}$ of $\mathrm{N}_{2}$ or $100 \mathrm{~g}$ of $\mathrm{NH}_{3}$ Solution: (a) Gram molar mass of $\mathrm{N}_{2}=2 \times 14=28 \mathrm{~g}$ $28 \mathrm{~g}$ of $\mathrm{N}_{2}$ have nitrogen atoms $=2 \times \mathrm{N}_{\mathrm{A}}$ $100 \mathrm{~g}$ of $\mathrm{N}_{2}$ have nitrogen atoms $=2 \times \mathrm{N}_{\mathrm{A}} \times \frac{(100 \mathrm{~g})}{(28 \mathrm{~g})}=7 \cdot 143 \times \mathrm{N}_{\mathrm{A}}$ $=7.143 \times 6.022 ...
Read More →Compute the difference in masses
Question: Compute the difference in masses of $10^{3}$ moles each of magnesium atoms and magnesium ions. Solution: $\mathrm{Mg}$ atom and $\mathrm{Mg}^{2+}$ ion differ by electrons $=2$ $10^{3}$ moles of $\mathrm{Mg}$ atoms and $\mathrm{Mg}^{2+}$ ions differ in electrons $=10^{3} \times 2$ moles Mass of one electron $=9.1 \times 10^{-31} \mathrm{~kg}$ Mass of $10^{3} \times 2$ moles of electrons $=10^{3} \times 2 \times 6.022 \times 10^{23} \times 9.1 \times 10^{-31} \mathrm{~kg}$ $=109.6 \times...
Read More →If A is 3 × 4 matrix and B is a matrix such that
Question: If $A$ is $3 \times 4$ matrix and $B$ is a matrix such that $A^{T} B$ and $B A^{T}$ are both defined. Then the order of $B$ is ______ Solution: Let $X=\left[x_{i j}\right]_{m \times n}$ and $Y=\left[y_{i j}\right]_{p \times q}$ be two matrices of order $m \times n$ and $p \times q$. The multiplication of matrices $X$ and $Y$ is defined if number of columns of $X$ is same as the number of rows of $Y$ i.e. $n=p$. Also, $X Y$ is a matrix of order $m \times q$. The order of matrix $A$ is $...
Read More →Using the formula, cos A
Question: Using the formula, $\cos A=\sqrt{\frac{1+\cos 2 A}{2}}$, find the value of $\cos 30^{\circ}$, it being given that $\cos 60^{\circ}=\frac{1}{2} .$ Solution: A= 30o⇒2A= 230o= 60o By substituting the value of the given T-ratio, we get: $\cos A=\sqrt{\frac{1+\cos 2 A}{2}}$ $\Rightarrow \cos 30^{\circ}=\sqrt{\frac{1+\cos 60^{\circ}}{2}}=\sqrt{\frac{1+\frac{1}{2}}{2}}=\sqrt{\frac{\frac{3}{2}}{2}}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$ $\therefore \cos 30^{\circ}=\frac{\sqrt{3}}{2}$...
Read More →Express each of the following in kilograms
Question: Express each of the following in kilograms (a) $5.84 \times 10^{-3} \mathrm{mg}$ (b) $58.34 \mathrm{~g}$ (c) $0.584 \mathrm{~g}$ (d) $5.873 \times 10^{-23} \mathrm{~g}$. Solution: (a) $5.84 \times 10^{-9} \mathrm{~kg}$ (b) $5.834 \times 10^{-4} \mathrm{~kg}$ (c) $0.584 \times 10^{-3} \mathrm{~kg}$ (d) $5.873 \times 10^{-26} \mathrm{~kg}$....
Read More →What is the SI prefix for each of the
Question: What is the SI prefix for each of the following multiples and submultiples of a unit? Solution: (a) kilo (b) deci (c) centri (d) micro (e) nano (f) pico...
Read More →Using the formula, tan 2 A
Question: Using the formula, $\tan 2 A=\frac{2 \tan A}{1-\tan ^{2} A}$, find the value of $\tan 60^{\circ}$, it being given that $\tan 30^{\circ}=\frac{1}{\sqrt{3}}$. Solution: $A=30^{\circ}$ $\Rightarrow 2 A=2 \times 30^{\circ}=60^{\circ}$ By substituting the value of the given T-ratio, we get: $\tan 2 A=\frac{2 \tan A}{1-\tan ^{2} A}$ $\Rightarrow \tan 60^{\circ}=\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=\frac{2 \times \frac{1}{\sqrt{3}}}{1-\left(\frac{1}{\sqrt{3}}\right)^{2}}=\frac{\fr...
Read More →The visible universe is estimated to contain
Question: The visible universe is estimated to contain $10^{22}$ stars. How many moles of stars are present in the visible universe? Solution: No. of stars present in visible universe $=10^{22}$ No. of moles of stars present $=\frac{10^{22}}{6.022 \times 10^{23}}=0.0166 \mathrm{~mol}$....
Read More →Solve this
Question: If $A=\left[\begin{array}{cc}\cos \theta \sin \theta \\ -\sin \theta \cos \theta\end{array}\right]$, then $A A^{T}=$ ________ Solution: The given matrix is $A=\left[\begin{array}{cc}\cos \theta \sin \theta \\ -\sin \theta \cos \theta\end{array}\right]$. $\therefore A A^{T}$ $=\left[\begin{array}{cc}\cos \theta \sin \theta \\ -\sin \theta \cos \theta\end{array}\right]\left[\begin{array}{cc}\cos \theta \sin \theta \\ -\sin \theta \cos \theta\end{array}\right]^{T}$ $=\left[\begin{array}{c...
Read More →Fill in the missing data in the given table
Question: Fill in the missing data in the given table Solution:...
Read More →If A and B are acute angles such that tan A
Question: If $A$ and $B$ are acute angles such that $\tan A=\frac{1}{3}, \tan B=\frac{1}{2}$ and $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$, show that $A+B=45^{\circ}$. Solution: Given: $\tan A=\frac{1}{3}$ and $\tan B=\frac{1}{2}$ $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$ On substituting these values in RHS of the expression, we get: $\frac{\tan A+\tan B}{1-\tan A \tan B}=\frac{\left(\frac{1}{3}+\frac{1}{2}\right)}{\left(1-\frac{1}{3} \times \frac{1}{2}\right)}=\frac{\left(\fra...
Read More →Raunak took 5 moles of carbon atoms in a container
Question: Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles of sodium atoms in another container of same weight. (a) Whose container is heavier? (b) Whose container has more number of atoms? Solution: (a) Molar atomic mass of carbon $=12 \mathrm{~g}$ No. of moles of carbon carried by Raunak $=5$ Mass of 5 moles of carbon $=(12 \times 5)=60 \mathrm{~g}$ Molar atomic mass of sodium $=23 \mathrm{~g}$ No. of moles of sodium carried by Krish $=5$ Mass of 5 moles of sodium...
Read More →A sample of vitamin C is known to contain
Question: A sample of vitamin $\mathrm{C}$ is known to contain $2.58 \times 10^{24}$ oxygen atoms. How many moles of oxygen atoms are present in the sample? Solution: $6.022 \times 10^{23}$ atoms of oxygen are present $=1 \mathrm{~mol}$ atoms $2.58 \times 10^{24}$ atoms of oxygen are present in $=\frac{(1 \mathrm{~mol} \text { atoms })}{\left(6.022 \times 10^{23}\right)} \times\left(2.58 \times 10^{24}\right)$ $=4.28 \mathrm{~mol}$ atoms $=4.28 \mathrm{~mol}$ atoms....
Read More →Solve the following equations
Question: If $A=\left[\begin{array}{cc}0 2 \\ 3 -4\end{array}\right]$ and $k A=\left[\begin{array}{cc}0 3 a \\ 2 b 24\end{array}\right]$, then $(k, a, b)=$_______ Solution: It is given that, $A=\left[\begin{array}{cc}0 2 \\ 3 -4\end{array}\right]$ and $k A=\left[\begin{array}{cc}0 3 a \\ 2 b 24\end{array}\right]$. $\therefore k\left[\begin{array}{cc}0 2 \\ 3 -4\end{array}\right]=\left[\begin{array}{cc}0 3 a \\ 2 b 24\end{array}\right]$ $\Rightarrow\left[\begin{array}{cc}0 2 k \\ 3 k -4 k\end{arr...
Read More →If A = 60° and B = 30°, verify that:
Question: IfA= 60 andB= 30, verify that: IfA= 60 andB= 30, verify that:(i) sin (AB) = sinAcosB cosAsinB(ii) cos (AB) = cosAcosB+ sinAsinB (iii) $\tan (\mathrm{A}-\mathrm{B})=\frac{\tan A-\tan B}{1+\tan A \tan B}$ Solution: (i) $\sin (A-B)=\sin 30^{\circ}=\frac{1}{2}$ sinAcosBcosAsinB= sin60ocos30ocos60osin30o $=\left(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}-\frac{1}{2} \times \frac{1}{2}\right)=\left(\frac{3}{4}-\frac{1}{4}\right)=\frac{2}{4}=\frac{1}{2}$ sin (AB) =sinAcosBcosAsinB (ii) $\co...
Read More →The mass of one steel screw is
Question: The mass of one steel screw is $4.11 \mathrm{~g}$. Find the mass of one mole of these steel screws. Compare this value with the mass of the Earth $\left(5.98 \times 10^{24} \mathrm{~kg}\right)$. Which one of the two is heavier and by how many times? Solution: No. of steel screws in 1 mole $=6.022 \times 10^{23}$ Mass of one steel screw $=4.11 \mathrm{~g}$ Mass of one mole steel screws $=(4.11 \mathrm{~g}) \times\left(6.022 \times 10^{23}\right)$ $=2.475 \times 10^{24} \mathrm{~g}=2.475...
Read More →Cinnabar (HgS) is a prominent ore of mercury.
Question: Cinnabar (HgS) is a prominent ore of mercury. How many grams of mercury are present in $225 \mathrm{~g}$ of pure HgS? Solution: Molar mass of cinnabar $(\mathrm{HgS})=(200 \cdot 6+32)=232 \cdot 6 \mathrm{~g} \mathrm{~mol}^{-1}$. Mass of mercury in $232.6 \mathrm{~g}$ of $\mathrm{HgS}=200.6 \mathrm{~g}$ Mass of mercury in $225 \mathrm{~g}$ of $\mathrm{HgS}=\frac{(200 \cdot 6 \mathrm{~g})}{(232 \cdot 6 \mathrm{~g})} \times(225 \mathrm{~g})=194 \cdot 04 \mathrm{~g}$....
Read More →The difference in the mass of 100 moles each of sodium
Question: The difference in the mass of 100 moles each of sodium atoms and sodium ions is $5.48002 \mathrm{~g}$. Compute the mass of an electron. Solution: Sodium $(\mathrm{Na})$ and sodium ion $\left(\mathrm{Na}^{+}\right)$differ in one electron. This means that 100 moles of $\mathrm{Na}$ atoms and $\mathrm{Na}^{+}$ ions differ in electrons $=100 \mathrm{~mol}$. No. of electrons in 100 moles $=100 \times 6.022 \times 10^{23}=6.022 \times 10^{25}$ Mass of $6.022 \times 10^{25}$ electrons $=5.480...
Read More →Calcium chloride when dissolved in water
Question: Calcium chloride when dissolved in water dissociates into its ions according to the following equation. $\mathrm{CaCl}_{2}(a q) \rightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q)$ Calculate the number of ions obtained from $\mathrm{CaCl}_{2}$ when $222 \mathrm{~g}$ of it is dissolved in water. Solution: Molar mass of $\mathrm{CaCl}_{2}=40+(2 \times 35 \cdot 5)=111 \mathrm{~g} \mathrm{~mol}^{-1}$ $111 \mathrm{~g}$ of $\mathrm{CaCl}_{2}$ represent $=1 \mathrm{~mol}$ $222 \mathrm{~g...
Read More →Solve the following equations
Question: If $e\left[\begin{array}{ll}e^{x} e^{y} \\ e^{y} e^{x}\end{array}\right]=\left[\begin{array}{ll}1 1 \\ 1 1\end{array}\right]$, then $x=$__________$y=$_________ Solution: $e\left[\begin{array}{ll}e^{x} e^{y} \\ e^{y} e^{x}\end{array}\right]=\left[\begin{array}{ll}1 1 \\ 1 1\end{array}\right]$ $\Rightarrow\left[\begin{array}{ll}e^{x} \times e e^{y} \times e \\ e^{y} \times e e^{x} \times e\end{array}\right]=\left[\begin{array}{ll}1 1 \\ 1 1\end{array}\right]$ $\Rightarrow\left[\begin{arr...
Read More →Find the ratio by mass of the combining elements in the following compounds:
Question: Find the ratio by mass of the combining elements in the following compounds: Solution:...
Read More →Verify by calculating that :
Question: Verify by calculating that : (a) 5 moles of $\mathrm{CO}_{2}$ and 5 moles of $\mathrm{H}_{2} \mathrm{O}$ do not have the same mass. (b) $240 \mathrm{~g}$ of calcium and $240 \mathrm{~g}$ magnesium elements have mole ratio of $3: 5$. Solution: (a) Mass of 5 moles of $\mathrm{CO}_{2}$ Mass of 1 mole of $\mathrm{CO}_{2}=12+2 \times 1=44 \mathrm{~g} \mathrm{~mol}^{-1}$. Mass of 5 moles of $\mathrm{CO}_{2}=(5 \mathrm{~mol}) \times\left(44 \mathrm{~g} \mathrm{~mol}^{-1}\right)=220 \mathrm{~g...
Read More →Calculate the number of moles of magnesium
Question: Calculate the number of moles of magnesium present in a magnesium ribon weighing $12 \mathrm{~g}$. Molar atomic mass of magnesium is $24 \mathrm{~g} \mathrm{~mol}^{-1}$. Solution: Mass of magnesium $(\mathrm{Mg})=12 \mathrm{~g}$ Molar atomic mass of magnesium $(\mathrm{Mg})=24 \mathrm{~g} \mathrm{~mol}^{-1}$. No. of moles of magnesium $(\mathrm{Mg})=\frac{(\text { Mass })}{(\text { Molar atomic mass })}=\frac{(12 \mathrm{~g})}{\left(24 \mathrm{~g} \mathrm{~mol}^{-1}\right)}=0.5 \mathrm...
Read More →If A = 60° and B = 30°, verify that:
Question: IfA= 60 andB= 30, verify that:(i) sin (A+B) = sinAcosB+ cosAsinB(ii) cos (A+B) = cosAcosB sinAsinB Solution: A= 60oandB=30oNow,A+B= 60o+30o = 90oAlso,AB=60o 30o= 30o(i) sin (A+B) = sin 90o= 1sinAcosB+cosAsinB= sin60ocos30o+cos60osin30o $=\left(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}+\frac{1}{2} \times \frac{1}{2}\right)=\left(\frac{3}{4}+\frac{1}{4}\right)=1$ sin (A+B) = sinAcosB+cosAsinB (ii) cos (A+B) = cos 90o = 0 $\cos A \cos B-\sin A \sin B=\cos 60^{\circ} \cos 30^{\circ}-\s...
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