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Chlorous acid ⇒ HClO?

Let P is a root   $\left(\in I\right)$

Case-I : p is odd

p¹⁰ + ap⁹ + b   $\ne 0$

Because LHS odd

Case-II : p is even

$p^{10}+a{p}^9+b\ne 0$  

because, LHs is odd

Initial m equivalent of Cu²? = 200 × 0.5 × 2 = 200 m eq
So electricity passed = (0.965×3600)/96500 = 36 × 10? ³ eq. = 36 m eq
m eq CuBr? remaining = 200 – 36 = 164 ∴ N = meq / V (in ml) = 164/200 = 0.82

A vector with zero magnitude is considered a zero vector. Since it has no magnitude, its direction is considered arbitrary. This means the zero vector has no specific direction.

Mol wt of propane = 44 g and weight of propane = 12.0 g
So, using PV = nRT
V = (nRT)/P = (12/44) * 0.082 * 333) / (740/760) = 7.648 L

From ΔT? = K? × m, and ΔT? = 15°C
∴ m = ΔT? / K? = 15 / 1.86 = 8.06
So the amount of propyl alcohol to be added.
= m × molwt = 8.06 × 60 = 483.6 g

The GATE 2026 exam will be conducted in more than 200 cities across eight zones in India. There are no international centres. Candidates who wish to appear for the exam must travel to India. The details regarding the test city and address will be mentioned in the GATE admit card. 

∴ % of S in the compound
= (32/233) × (mass of BaSO? / mass of compound) × 100 = (32 × 0.35 × 100) / (233 × 0.25) = 19.227 ≈ 19.23

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Ans. (8.00)

Hi, Power BI can be learned by following the steps mentioned in the table below. I have tried to keep it simple and in layman's language for the absolute beginners.

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Factual on lyophilic colloids
(A) It is easy to prepare
(B) Stable
(C) Revisable
(D) Viscosity is high and surface tension is low for DP. and DM.

Factual
(A) Magnalium contain Al and Mg
(B) Bell metal ⇒ Cu and Sn
(C) Gun metal ⇒ Cu, Sn and Zn
(D) Chrome steel ⇒ Fe and Cr.

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GATE enrollment ID is created at time of registration. It is a login credential. GATE registration is roll number given to students. The registration number is printed on hall ticket.

Sm ⇒ +2, +3 (4f? , 6 s²); Gd ⇒ +3 (4f?5 d¹6 s²)
Tm ⇒ +2 + 3 (4f¹³,6 s²); Nd ⇒ +2, +3, +4 (4f? , 6 s²)

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Ans. (8.40 to 8.57)

${\eta }_1=1-\frac{T_2}{T_1},{\eta }_2=1-\frac{T_3}{T_2}$

as ${\eta }_1={\eta }_2\Rightarrow \frac{T_2}{T_1}=\frac{T_3}{T_2}\Rightarrow T_2={\left(T_1{T}_3\right)}^{\frac{1}{2}}=\sqrt[]{400\times 900}=600\mathrm{K}$

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Ans. (122.00)

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