True or false solar power take power from wind and uses it to make electricity

Potassium thiocyanate, KSCN, is often used to detect the presence of Fe3+ ions in solution through the formation of the red Fe(H

Natalija [7]

Answer:

Ferric ions left in the solution at equilibrium is $8.0\times 10^{-6} M$ .

Explanation:

Moles of ferric nitrate in 0.700 L = n

Volume of the solution = V = 0.700 L

Molarity of ferric nitrate = M = 0.00150 M

$n=M\times V=0.00150 M\times 0.700 L=0.00105 mol$

Moles of potassium thiocyanate in 0.700 L = n'

Volume of the potassium thiocyanate solution = V' = 0.700 L

Molarity of potassium thiocyanate = M' = 0.200 M

$n'=M'\times V'=0.200 M\times 0.700 L=0.140 mol$

Molarity of ferric ions after mixing :

1 mol of ferric nitrate gives 1 mol of ferric ions.Then 0.00105 mol ferric nitrate will :

Moles of ferric ions = 0.00105 mol

$M_1=\frac{0.00105 mol}{0.700 L+0.700 L}=0.00075 M$

Molarity of thiocyanate ions after mixing :

1 mol of potassium thiocyanate gives 1 mol of thiocyanate ions.Then 0.140 mol potassium thiocyanate will give:

Moles of thiocyanate ions = 0.140 mol

$M_2=\frac{0.140 mol}{0.700 L+0.700 L}=0.1 M$

Complex equation:

$Fe^{3+}+SCN^-\rightleftharpoons [Fe(SCN)]^{2+}$

0.00075 M 0.1 M 0

At equilibrium:

(0.00075 M -x) (0.1 M-x) x

The formation constant of the given complex = $K_f=8.9\times 10^2$

$K_f=\frac{[[Fe(SCN)]^{2+}]}{[[Fe^{3+}]][SCN^{-}]}$

$8.9\times 10^2=\frac{x}{(0.00075 M -x)\times (0.1 M-x)}$

Solving for x:

x = 0.000742 M

Ferric ions left in the solution at equilibrium :

= (0.00075 M -x) = (0.00075 M - 0.000742 M)= $8.0\times 10^{-6} M$

5 0

2 years ago

Sulfurous acid, H2SO3, breaks down into water (H20) and sulfur dioxide (SO2).

erastovalidia [21]

Answer:

1 atom of sulfur

Explanation:

5 0

3 years ago

A certain drug has a half-life in the body of 3.5h. Suppose a patient takes one 200.Mg pill at :500PM and another identical pill

tekilochka [14]

Answer:

The amount of drug left in his body at 7:00 pm is 315.7 mg.

Explanation:

First, we need to find the amount of drug in the body at 90 min by using the exponential decay equation:

$N_{t} = N_{0}e^{-\lambda t}$

Where:

λ: is the decay constant = $ln(2)/t_{1/2}$

$t_{1/2}$ : is the half-life of the drug = 3.5 h

N(t): is the quantity of the drug at time t

N₀: is the initial quantity

After 90 min and before he takes the other 200 mg pill, we have:

$N_{t} = 200e^{-\frac{ln(2)}{3.5 h}*90 min*\frac{1 h}{60 min}} = 148.6 mg$

Now, at 7:00 pm we have:

$t = 7:00 pm - (5:00 pm + 90 min) = 30 min$

$N_{t} = (200 + 148.6)e^{-\frac{ln(2)}{3.5 h}*30 min*\frac{1 h}{60 min}} = 315.7 mg$

Therefore, the amount of drug left in his body at 7:00 pm is 315.7 mg (from an initial amount of 400 mg).

I hope it helps you!

3 0

2 years ago

How do I cook eggs? Please tell me

svetoff [14.1K]

You just throw them on the pan with the flame all the way up it will be done in a second !

3 0

2 years ago

Read 2 more answers

How many grams are in a sample of 0.55 mol of K?

Katyanochek1 [597]

The equation you use here is
mass =moles x Mr

So:
Moles of K - 0.55mol
Mr of K - 39.1

Mass= 0.55x39.1 =21.505g

4 0

2 years ago