0-1-1-3-1000 - h-800 - 0-1 - h-1000

0.1×(1.3× 1000)=h× 800+(0.1 h)× 1000 0.1× 1300=h× 800+0.1× 1000 h× 1000 130=h× 800+100 h× 1000 130 100=h× 800 h× 1000 30=h× (800 1000) ...

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Byju's Answer

Standard X

Chemistry

Relationship between Vapor Density & Relative Molecular Mass

0.1×1.3×1000 ...

Question

$0 .1 \times (1 .3 \times 1000) = h \times 800 + (0 .1 - h) \times 1000$

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Solution

$0.1 \times (1.3 \times 1000) = h \times 800 + (0.1 - h) \times 1000$
$0.1 \times 1300 = h \times 800 + 0.1 \times 1000 - h \times 1000$
$130 = h \times 800 + 100 - h \times 1000$
$130 - 100 = h \times 800 - h \times 1000$
$30 = h \times (800 - 1000)$
$30 = h \times (- 200)$
$\frac{30}{- 200} = h$

$- 0.15 = h$

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Similar questions

In the given figure, the load weighs 200N, $θ$ = $37^{\circ}$ and beam has negligible mass. Then, the tension in string and horizontal and vertical forces that pivot exerts on the beam is

Assertion (A) :

[\frac{1}{2}] + [\frac{1}{2} + \frac{1}{1000}] + [\frac{1}{2} + \frac{2}{1000}] + [\frac{1}{2} + \frac{3}{1000}] + . . . + [\frac{1}{2} + \frac{999}{1000}] = 500

(where [.] denotes G.I.F.)

Reason (R) :

[\frac{1}{2} + \frac{r}{1000}] = {\begin{matrix} 0; & if 0 \leq r < 500 1; & if 500 \leq r \leq 999 \end{matrix}

Q. The solution set of the inequality

({log}_{10} x)^{4} - 13 ({log}_{10} x)^{2} + 36 > 0

Complete the following:
(a) $1 k W h = \frac{1 v o l t \times 1 a m p e r e \times \dots}{1000}$
(b) $1 k W h = \dots j$

lim h \to 0 [\frac{1}{h \sqrt[3]{8 + h}} - \frac{1}{2 h}]

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0.1×(1.3×1000)=h×800+(0.1−h)×1000

0.1×(1.3×1000)=h×800+(0.1−h)×10000.1×1300=h×800+0.1×1000−h×1000130=h×800+100−h×1000130−100=h×800−h×100030=h×(800−1000)30=h×(−200)30−200=h−0.15=h

$0 .1 \times (1 .3 \times 1000) = h \times 800 + (0 .1 - h) \times 1000$