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ira [324]
4 years ago
15

Find the time Δt it takes the magnetic field to drop to zero. Express your answer in terms of some or all of the quantities a, B

0, I, N, and R.
Chemistry
1 answer:
Mekhanik [1.2K]4 years ago
7 0
Δt= \frac{NB_0 \pi a^2 }{IR}

This because ε = dΦ/dt, and emf is a voltage so it can also be written as dΦ/dt = IR

Because ΔΦ= ABcosθ for each of the N loops and cosθ is just 1 here only A needs to be replaced, and it can be replaced with A=πr^2, which makes that part of the equation ΔΦ_m=N*B0*πa^2
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How many grams of a stock solution that is 92.5 percent H2SO4 by mass would be needed to make 250 grams of a 35.0 percent by mas
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M1m1 = M2m2

where M1 is the concentration of the stock solution, m1 is the mass of the stock solution, M2 is the concentration of the new solution and m2 is its new mass.

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Hydrogen is manufactured on an industrial scale by this sequence of reactions: Write an equation that gives the overall equilibr
RideAnS [48]

The question is incomplete. The complete question is :

Hydrogen is manufactured on an industrial scale by this sequence of reactions:

$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g    ) \ \ \ \ \ \ \ \ \ \ K_1$

$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \ \ \ \ \ \ \ \ \ \ \ \  K_2$

The net reaction is  :

$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g) \ \ \ \ \ \ \ \ \ K$

Write an equation that gives the overall equilibrium constant K in terms of the equilibrium constants K_1 and K_2. If you need to include any physical constants, be sure you use their standard symbols, which you'll find in the ALEKS Calculator.

Solution :

$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g    ) \ \ \ \ \ \ \ \ \ \ K_1$

$K_1 = \frac{[CO][H_2]^3}{[CH_4][H_2O]}$     ...............(1)

$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \ \ \ \ \ \ \ \ \ \ \ \  K_2$

$K_2 = \frac{[CO_2][H_2]}{[CO][H_2O]}$  ...................(2)

$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g) \ \ \ \ \ \ \ \ \ K$

$K=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$

On multiplication of equation (1) and (2), we get

$K_1 \times K_2=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \times \frac{[CO_2][H_2]}{[CO][H_2O]}$

$K_1K_2=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$  .................(4)

Comparing equation (3) and equation (4), we get

$K=K_1K_2$

4 0
3 years ago
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