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1 year ago

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1 Ten (10) ml aqueous solutions of drug A (10% w/v) and drug B (25% w/v) are stored in two identical test tubes under identical

storage conditions at 37°C for 3 months. If both drugs degrade by first-order, which drug will retain the highest percentage of initial concentration?

1 answer:

Reil [10]1 year ago

3 0

Answer:

YOUR answer is given below:

Explanation:

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The electric potential in a region that is within 2.00 mm of the origin of a rectangular coordinate system is given by V=Axl+Bym

Fynjy0 [20]

Answer:

Given the potential, $V = Ax^l+By^m+Cz^n+D$

The components of the electric field are:

$E_x = \frac{-dV}{dx} = -Alx^l^-^1$

$E_y = \frac{-dV}{dy} = - Bmy^m^-^1$

$E_z = \frac{-dV}{dz} = - nCzn^n^-^1$

Let's calculate the potential difference for all given points.

$V(0, 0, 0) = 10V => Ax^l+By^m+Cz^n+D = 10$

$=> D = 10$

$V(1, 0, 0) = 4V => A + 10 = 4$

Solving for A, we have:

$A = 4 - 10$

$A = -6$

$V(0, 1, 0) = 6V => B + 10 = 6$

Solving for B, we have:

$B = 6 - 10$

$B = -4$

$V(0, 0, 1) = 8V => C + 10 = 4$

Solving for C, we have:

$C = 8 - 10$

$C = -2$

For all given points, let's calculate the magnitude of electric field as follow:

$E_x (1, 0, 0) = 16 => - Alx^l^-^1 = 16$

$Al = -16$

Solving for l, we have:

$l = \frac{-16}{A}$

From above, A = -6

$l = \frac{-16}{-6}$

$l = \frac{8}{3}$

$E_y (0, 1, 0) = 16=> Bmy^m^-^1 = 16$

$Bm = -16$

Solving for m, we have:

$m = \frac{-16}{A}$

From above, B = -4

$m = \frac{-16}{-4}$

$m = 4$

$E_y (0, 0, 1) = 16=> nCz^n^-^1 = 16$

$nC = - 16$

Solving for n, we have:

$n = \frac{-16}{C}$

From above, C = -2

$n = \frac{-16}{-2}$

$n = 8$

4 0

1 year ago

A charge of 5.67 x 10-18 C is placed 3.5 x 10 m away from another charge of - 3.79 x 10 "C

miskamm [114]

Answer:

1. 579 x 10 ^-22N

Explanation:

F = kq1q2/r^2

= 9.0 x 10^9 x 5.67 x 10^-18 x 3.79 x 10^-18/ (3.5 x 10^-2)^2

= 1. 579 x 10 ^-22N

6 0

2 years ago

Um navio cargueiro proveniente do Oceano Atlântico passa a navegar nas águas menos densas do rio Amazonas. Em comparação com a s

LUCKY_DIMON [66]

Answer:

Mas eu acho q é o empuxo será igual e a porção imersa do navio será maior.

Explanation: SE

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8 0

2 years ago

Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i

SCORPION-xisa [38]

Answer:

The value is $r = 5.077 \ m$

Explanation:

From the question we are told that

The Coulomb constant is $k = 9.0 *10^{9} \ N\cdot m^2 /C^2$

The charge on the electron/proton is $e = 1.6*10^{-19} \ C$

The mass of proton $m_{proton} = 1.67*10^{-27} \ kg$

The mass of electron is $m_{electron } = 9.11 *10^{-31} \ kg$

Generally for the electron to be held up by the force gravity

Then

Electric force on the electron = The gravitational Force

i.e

$m_{electron} * g = \frac{ k * e^2 }{r^2 }$

$\frac{9*10^9 * (1.60 *10^{-19})^2 }{r^2 } = 9.11 *10^{-31 } * 9.81$

$r = \sqrt{25.78}$

$r = 5.077 \ m$

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2 years ago

In the ENGR 10 lab (E391), there are 50 long light bulbs (P=100 W) and 30 regular bulbs (P=60 W). How much energy is consumed li

Alenkinab [10]

Answer:

Total energy saving will be 0.8 KWH

Explanation:

We have given there are 50 long light bulbs of power 100 W so total power of 50 bulb = 100×50 = 5000 W = 5 KW

30 bulbs are of power 60 W

So total power of 30 bulbs = 30×60 = 1800 W = 1.8 KW

Total power of 80 bulbs = 1.8+5 = 6.8 KW

Total time = 3 hour

We know that energy $E=power\times time=6.8\times 3=20.4KWH$

Now power of each CFL bulb = 25 W

So power of 80 bulbs = 80×25 = 2000 W = 2 KW

Energy of 80 bulbs = 2×3 = 6 KWH

So total energy saving = 6.8-6 = 0.8 KWH

6 0

2 years ago