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4 years ago

A dioxin-contaminated water source contains 0.085% dioxin by mass. how much dioxin is present in 2.5 l of this water?

1 answer:

8 0

The mass of dioxin present in 2.5L of water is 2.125 grams

calculation
convert the 2.5 l to Ml = 2.5 x1000 = 2500 Ml

density of water is = 1g/ml

mass is of water is therefore = density x volume

= 2500 Ml x 1g/ml = 2500 grams
by of the below formula of calculating of % mass

that is mass of the solute / mass of the solution = % mass

let the mass of solute be represented by y
mass of solution = 2500 g
% mass = 0.085% = 0.085/100 into fraction

by subsituting in the formula

=y/2500= 0.085/100
cross multiplication

100y =212.5

divide both side by 100

y= 2.125 grams

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Sheila's measured glucose level one hour after a sugary drink varies according to the normal distribution with μμ = 120 mg/dl an

Molodets [167]

Since the sample size is below 30, in this case we use the t statistic. The formula for t score is:

t = (x – u) / (σ / sqrt n)

where,

x = the level l = unknown

u = sample mean = 120 mg / dl

σ = standard deviation = 20 mg / dl

n = sample size or number of results = 5

Using the standard distribution tables for t, we can find the value of t given the probability (P = 0.15) and degrees of freedom (DOF).

t = 1.036

Going back to the formula for t score:

1.036 = (x – 120) / (20 / sqrt 5)

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8 0

3 years ago

Consider the two gaseous equilibria: The values of the equilibrium constants K 1 and K 2 are related by

Sauron [17]

The question is missing parts. The complete question is as follows.

Consider the two gaseous equilibria involving SO2 and the corresponding equilibrium constants at 298K:

$SO_{2}_{(g)} + \frac{1}{2}O_{2}$ ⇔ $SO_{3}_{(g)}$ ; $K_{1}$

$2SO_{3}_{(g)}$ ⇔ $2SO_{2}_{(g)}+O_{2}_{(g)}; K_{2}$

The values of the equilibrium constants are related by:

a) $K_{1}$ = $K_{2}$

b) $K_{2} = K_{1}^{2}$

c) $K_{2} = \frac{1}{K_{1}^{2}}$

d) $K_{2}=\frac{1}{K_{1}}$

Answer: c) $K_{2} = \frac{1}{K_{1}^{2}}$

Explanation: Equilibrium constant is a value in which the rate of the reaction going towards the right is the same rate as the reaction going towards the left. It is represented by letter K and is calculated as:

$K=\frac{[products]^{n}}{[reagents]^{m}}$

The concentration of each product divided by the concentration of each reagent. The indices, m and n, represent the coefficient of each product and each reagent.

The equilibrium constants of each reaction are:

$SO_{2}_{(g)} + \frac{1}{2}O_{2}$ ⇔ $SO_{3}_{(g)}$

$K_{1}=\frac{[SO_{3}]}{[SO_{2}][O_{2}]^{1/2}}$

$2SO_{3}_{(g)}$ ⇔ $2SO_{2}_{(g)}+O_{2}_{(g)}$

$K_{2}=\frac{[SO_{2}]^{2}[O_{2}]}{[SO_{3}]^{2}}$

Now, analysing each constant, it is easy to see that $K_{1}$ is the inverse of $K_{2}$ .

If you doubled the first reaction, it will have the same coefficients of the second reaction. Since coefficients are "transformed" in power for the constant, the relationship is:

$K_{2}=\frac{1}{K_{1}^{2}}$