Five darts strike near the center of the target. Who ever threw the darks is?

A standard solution of FeSCN2+ is prepared by combining 9.0 mL of 0.20 M Fe(NO3)3 with 1.0 mL of 0.0020 M KSCN . The standard so

Xelga [282]

Answer : The equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

Explanation :

First we have to calculate the initial moles of $Fe^{3+}$ and $SCN^-$ .

$\text{Moles of }Fe^{3+}=\text{Concentration of }Fe^{3+}\times \text{Volume of solution}$

$\text{Moles of }Fe^{3+}=0.20M\times 9.0mL=1.8mmol$

and,

$\text{Moles of }SCN^-=\text{Concentration of }SCN^-\times \text{Volume of solution}$

$\text{Moles of }SCN^-=0.0020M\times 1.0mL=0.0020mmol$

The given balanced chemical reaction is,

$Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)$

Since 1 mole of $Fe^{3+}$ reacts with 1 mole of $SCN^-$ to give 1 mole of $FeSCN^{2+}$

The limiting reagent is, $SCN^-$

So, the number of moles of $FeSCN^{2+}$ = 0.0020 mmole

Now we have to calculate the concentration of $FeSCN^{2+}$ .

$\text{Concentration of }FeSCN^{2+}=\frac{0.0020mmol}{9.0mL+1.0mL}=0.00020M$

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

where,

A = absorbance of solution

C = concentration of solution

l = path length

$\epsilon$ = molar absorptivity coefficient

$\epsilon$ and l are same for stock solution and dilute solution. So,

$\epsilon l=\frac{A}{C}=\frac{0.480}{0.00020M}=2400M^{-1}$

For trial solution:

The equilibrium concentration of $SCN^-$ is,

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{initial}$ = 0.00050 M

Now calculate the $[FeSCN^{2+}]$ .

$C=\frac{A}{\epsilon l}=\frac{0.220}{2400M^{-1}}=9.17\times 10^{-5}M$

Now calculate the concentration of $SCN^-$ .

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{eqm}=(0.00050M)-(9.17\times 10^{-5}M)$

$[SCN^-]_{eqm}=4.58\times 10^{-8}M$

Therefore, the equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

5 0

3 years ago

How long does it take the sun to melt a block of ice at 0∘c with a flat horizontal area 1.0 m2 and thickness 1.8 cm ? assume tha

VLD [36.1K]

This is a problem involving heat transfer through radiation. The solution to this problem would be to use the formula for heat flux.

ΔQ/Δt = (1000 W/m²)∈Acosθ

A is the total surface area:
A = (1 m²) + 4(1.8 cm)(1m/100 cm)(√(1 m²))
A = 1.072 m²

ΔQ is the heat of melting ice.
ΔQ = mΔHfus
Let's find its mass knowing that the density of ice is 916.7 kg/m³.
ΔQ = (916.7 kg/m³)(1 m²)(1.8 cm)(1m/100 cm)(<span>333,550 J/kg)
</span>ΔQ = 5,503,780 J

5,503,780 J/Δt = (1000 W/m²)(0.05)(1.072 m²)(cos 33°)
<em>Δt = 122,434.691 s or 34 hours</em>

4 0

3 years ago

Questions on maths mechanics

Sav [38]

Answer:

I don't understand your questions

3 0

3 years ago

For each value of the principal quantum number n, what are the possible values of the electron spin quantum number?

Ede4ka [16]

Answer:

The values for spin quantum number +1/2 and - 1/2

Explanation:

Principal quantum number denoted by (n) is used to describe the shell or orbits that electrons are found. Principal quantum number can assume a value of n= 1,2, 3, 4,5............ which indicates K, L, M, N, O shell respectively.

To know the maximum number of electrons in each shell, the formula (2n²) can be used. The letter 'n' denotes the values of principal quantum number 1,2,3,4

For example

n=1 (K shell) has maximum number of 2 electrons
n=2 (L shell) has the maximum number of 8 electrons
n=3 (M shell) has the maximum number of 18 electrons
n=4 (N shell) has the maximum number of 32 electrons

All the electron in each shell will have a spin quantum number of +1/2 and - 1/2. One electron in each degenerate orbital will spin up (+1/2) while the other electron will spin down (-1/2).

7 0

3 years ago

How can an object enter Earth's orbit?

pentagon [3]

When an object enters the Earth's atmosphere, it experiences a few forces, including gravity and drag. Gravity will naturally pull an object back to earth.

-Hope this helped, have a great day. ;)

3 0

2 years ago