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3 years ago

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What are differences and similarities between pure substances and mixtures

1 answer:

Snezhnost [94]3 years ago

5 0

Answer:

HELLO THERE!

I HOPE MY ANSWER WILL HELP YOU :)

Explanation:

There is this picture that helped me. I hope it helps you too.

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A. Based on the activation energies and frequency factors, rank the following reactions from fastest to slowest reaction rate, a

deff fn [24]

Answer:

A) $E_{a} = 350KJ/mol$ , $E_{a} = 50KJ/mol$ , $E_{a} = 50KJ/mol$

A = 1.5× $10^{-7}s^{-1}$ , A = 1.9× $10^{-7} s^{-1}$ , A=1.5× $10^{-7} s^{-1}$

B) 4.469

Explanation:

From Arrhenius equation

$K=Ae^{\frac{E_{a} }{RT} }$

where; K = Rate of constant

A = Pre exponetial factor

$E_{a}$ = Activation Energy

R = Universal constant

T = Temperature in Kelvin

Given parameters:

$E_{a} =165KJ/mol$

$T_{1}=505K$

$T_{2}=525K$

$R=8.314JK^{-1}mol^{-1}$

taking logarithm on both sides of the equation we have;

$InK=InA-\frac{E_{a} }{RT}$

since we have the rate of two different temperature the equation can be derived as:

$In(\frac{K_{2} }{K_{1} } )=\frac{E_{a} }{R}(\frac{1}{T_{1} } -\frac{1}{T_{2} } )$

$In(\frac{K_{2} }{K_{1} } )=\frac{165000J/mol}{8.314JK^{-1}mol^{-1} }.(\frac{1}{505} -\frac{1}{525} )$

$In(\frac{K_{2} }{K_{1} } )$ = 19846.04×7.544× $10^{-5}$ = 1.497

$\frac{K_{2} }{K_{1} }$ = $e^{1.497}$ = 4.469

6 0

3 years ago

Identify two factors that can produce metamorphic rocks

alexdok [17]

Heat & pressure. hope this helps

8 0

3 years ago

Read 2 more answers

I bet no one can solve this

Nikitich [7]

The answer is 89.4 g/mol cuz the two units carry positives so that means 8.50 flows for 2 more hours than usual.

3 0

3 years ago

Choose all the answers that apply.

zmey [24]

Calcium is a non metal

5 0

3 years ago

Read 2 more answers

A chemist adds 485 mL of a 0.0025 mol/L calcium sulfate solution to a reaction flask. Calculate the mass in grams of calcium sul

viva [34]

Answer : The mass in grams of calcium sulfate is 0.16 grams.

Explanation :

Molarity : It is defined as the number of moles of solute present in one litre of solution.

Formula used :

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution}}$

Solute is, $CaSO_4$

Given:

Molarity of $CaSO_4$ = 0.0025 mol/L

Molar mass of $CaSO_4$ = 136 g/mole

Volume of solution = 485 mL

Now put all the given values in the above formula, we get:

$0.0025=\frac{\text{Mass of }CaSO_4\times 1000}{136\times 485}$

$\text{Mass of }CaSO_4=0.16g$

Thus, the mass in grams of calcium sulfate is 0.16 grams.

6 0

3 years ago