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3 years ago

Please help me please

Chemistry

1 answer:

Harlamova29_29 [7]3 years ago

3 0

Answer:

Sugar solution seperation method is Distillation

iron power and sand separation method is Magnetic separation

petrol and diesel separation method is Fractional Distillation

camphor and galss powder seperation method is Sublimation

have a great day ❤️

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Consider the reaction data.

Snowcat [4.5K]

Answer:

(a) x₁= 0.004 444; (b) y₁ = -0.9545; (c) x₂ = 0.001 905; (d) y₂ = -0.4541;

(e) rise = 0.5004; (f) run = -0.002 539; (g) slope = -197.1; (h) Eₐ = -1.64 kJ·mol⁻¹

Explanation:

This is an example of the Arrhenius equation:

$k = Ae^{-E_{a}/RT}\\\text{Take the ln of each side}\\\ln k = \ln A - \dfrac{E_{a}}{RT}\\\\\text{We can rearrange this to give}\\\ln k = -\left ( \dfrac{E_{a}}{R} \right )\dfrac{1}{T} + \ln A$

Thus, if you plot ln k vs 1/T, you should get a straight line with slope = -Eₐ/R and a y-intercept = lnA

(a) x₁

x₁= 1/T₁ = 1/225 = 0.004 444

(b) y₁

y₁ = ln(k₁) = ln0.385 = -0.9545

(c) x₂

x₂= 1/T₂ = 1/525 = 0.001 905

(d) y₂

y₂ = ln(k₂) = ln0.635 = -0.4541

(e) Rise

Δy = y₂ - y₁ = -0.4541 - (-0.9545) = -0.4541 + 0.9545 = 0.5004

(f) Run

Δx = x₂ - x₁ = 0.001 905 - 0.004 444 = -0.002 539

(g) Slope

Δy/Δx = 0.5004/(-0.002 539) K⁻¹ = -197.1

(h) Activation energy

Slope = -Eₐ/R

Eₐ = -R × slope = -8.314 J·K⁻¹mol⁻¹ × (-197.1 K⁻¹) = 1638 J/mol = 1.64 kJ/mol

3 0

3 years ago

Please help me, It's my last question :)

Vedmedyk [2.9K]

<h3>Given:</h3>

$V_{1} = \text{5 L}$

$T_{1} = \text{298 K}$

$V_{2} = \text{3 L}$

<h3>Unknown:</h3>

$T_{2}$

<h3>Solution:</h3>

$\dfrac{V_{1}}{T_{1}} = \dfrac{V_{2}}{T_{2}}$

$T_{2} = T_{1} \times \dfrac{V_{2}}{V_{1}}$

$T_{2} = \text{298 K} \times \dfrac{\text{3 L}}{\text{5 L}}$

$\boxed{T_{2} = \text{179 K}}$

$\\$

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3 0

3 years ago

14. All the following are necessary parts of a neutralization reaction except A. an indicator. B. a salt. C. water. D. an acid.

ValentinkaMS [17]

The correct answer is <em>B. a Salt </em><em>because The reaction of an acid and a base is called a neutralization reaction because the properties of both the acid and base are diminished or neutralized when they react. A neutralization reaction is a reaction of an acid with a base in aqueous solution to produce water and a salt, as shown by the following equation:</em>

<em>acid + base → salt + water</em>

<em />

<em>* Hopefully this helps:) Mark me the brainliest:) </em>

6 0

3 years ago

Two concentration cells are prepared, both with 90.0 mL of 0.0100 M Cu(NO₃)₂ and a Cu bar in each half-cell. (b) Calculate Ecell

pogonyaev

The Ecell when an additional 10.0 mL of 0.500 M NH₃ is added is 0.156 V.

When NH3 is added to the first cell, Nh3 react with Cu(NO3) react to form complex.

Thus, Cu2+ ion concentration decrease in the first cell.

Anode

Cu ---- Cu(2+) + 2e-

Cathode

Cu(2+) + 2e- ------ Cu

Ecell can be calculated as

Ecell = E°cell - (0.059/2) log{ Cu(2+) / Cu(+2) cathode}

[Cu2+] cathode = 90ml × 0.01M = 9 × 10^(-4) moles

or,

0.129 = 0 - (0.059/2) log ( Cu(+2) / 9 × 10^(-4))

[Cu(2+) ] anode = 3.8 × 10^(-8) mol

<h3>Chemical reaction of Nh3 with Cu2+</h3>

(Cu2+) + 4 NH3 -----; Cu(NH3)4(2+)

Kf can be given as

Kf = [Cu(NH3)4(2+)]/ [Cu2+] [ NH3]^4

Concentration of NH3 = 19 ml × 0.5 M

= 0.005 m

Kf = 0.005/ (3.8 × 10^(-8) mol) × (0.005) ^4

= 2.09 × 10^14.

If 10ml NH3 id added in the solution, then the total concentration of NH3 can be 20ml and 0.5 M = 0.01mol

Now, we can calculate the [Cu2+] anode

[Cu2+] anode = [Cu(NH3)4(2+)]/ Kf × [ NH3]^4

By substituting all the values, we get

= 4.78 × 10^(-9) moles.

E cell = E°cell - (0.059/2) log{ Cu(2+) / Cu(+2)

0- (0.059/2) log{ 4.78 × 10^(-9) / 9 × 10^(-4))

E cell = 0.156 V.

Thus, we calculated that the Ecell when an additional 10.0 mL of 0.500 M NH₃ is added is 0.156 V.

learn more about Ecell:

brainly.com/question/861659

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7 0

2 years ago

PLEASE HELP!!!<br><br><br>Calculate the PH of an acid solution containing 0.10 M HNO3

Anika [276]

Answer:

The pH would be 1.

Explanation:

The pH of an acid can be found by using the formula pH = - log [H⁺]. This was a method developed to be able to measure the pH of substances.

Since the concentration of HNO₃ is 0.1 M, then it also means that the concentration of H⁺ ions is 0.1 M.

By substituting for the formula: pH of 0.1 M HNO₃ = - log [0.1]

= 1

3 0

3 years ago

Please help me please ​

Please help me please