Class 11th
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New answer posted
2 months agoContributor-Level 9
Oxidation state of N
| NO | +2 |
|-|-|
| NO? | +4 |
| N? O | +1 |
| NO? | +5 |
So, order of oxidation state is
NO? > NO? > NO > N? O
New answer posted
2 months agoContributor-Level 10
The data consists of n values of a and n values of -a.
Mean x? = (n*a + n* (-a) / 2n = 0 / 2n = 0.
Variance σ² = (Σx? ²)/N - x? ² = (n*a² + n* (-a)²) / 2n - 0² = 2na² / 2n = a².
If a value b is added to all observations, the new mean is x? ' = x? + b = 0 + b = b.
We are given the new mean is 5, so b=5.
Adding a constant does not change the variance. The new variance is still a².
We are given the new standard deviation is 20, so the new variance is 20² = 400.
Thus, a² = 400.
The required value is a² + b² = 400 + 5² = 400 + 25 = 425.
New answer posted
2 months ago
Contributor-Level 9
Hybridisation of carbon a, b, and c respectively are sp³, sp² and sp².
New answer posted
2 months agoContributor-Level 10
The equation is for a hyperbola: x²/4 - y²/2 = 1.
The eccentricity e is given by e = √ (1 + b²/a²) = √ (1 + 2/4) = √6/2.
The focus F is at (ae, 0), which is (2 * √6/2, 0) = (√6, 0).
The equation of the tangent at a point P (x? , y? ) is xx? /a² - yy? /b² = 1.
The equation of the tangent at P is given as x - (√6 y)/2 = 1.
This tangent cuts the x-axis (y=0) at x=1, so Q is (1, 0).
The latus rectum is the line x = ae = √6.
To find the point R where the tangent intersects the latus rectum, we substitute x=√6 into the tangent equation:
√6 - (√6 y)/2 = 1 ⇒ √6 - 1 = (√6 y)/2 ⇒ y = 2 (√6 - 1)/√6.
The vertices o
New answer posted
2 months agoContributor-Level 10
Given a right-angled triangle with one angle π/3 (or 60°), and the side adjacent to it is h/2.
Using the tangent function: tan (π/3) = opposite / adjacent.
Let the adjacent side be 2. Then tan (π/3) = h/2.
We know tan (π/3) = √3.
So, h/2 = √3, which implies h = 2√3.
New answer posted
2 months agoContributor-Level 10
Given a triangle with inradius r and circumradius R.
The inradius r is calculated as r = |0 + 0 - 3| / √2 = 3 / √2.
From the geometry of the triangle, we have the relation r/R = sin (30°) = 1/2.
This gives R = 2r.
The question asks for the sum R + r, which is 2r + r = 3r.
Substituting the value of r, we get 3 * (3/√2) = 9/√2.
∴ R + r = 9/√2.
New answer posted
2 months agoContributor-Level 10
(1/2)kx² = (1/2)mv² ⇒ x = v√ (m/k) = 10 * √ (4/100) = 10 * (2/10) = 2m
New answer posted
2 months agoContributor-Level 9
Mn? O? is mixed oxide which contains MnO and Mn? O? , so Mn is in +2 and +3 oxidation state respectively. Both Mn? ² & Mn³? has unpaired electrons so Mn? O? will show magnetic property. While all other oxides have no unpaired electrons either on cation or on anion.
New answer posted
2 months ago
Contributor-Level 10
|Δp| = 2mu sinθ = 2 * 5 * 10? ³ * 5√2 * (1/√2) = 5 * 10? ² kg m s? ¹
⇒ x = 5
New answer posted
2 months agoContributor-Level 10
V = (4/3)πr³
⇒ ΔV/V = 3 (Δr/r) ⇒ Relative error
% error in volume = (ΔV/V) * 100 = 3 (Δr/r) * 100 = 3 * (0.85/7.50) * 100 = 33.999 ≈ 34
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