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2 years ago

Can somebody give me a example of a mixture?

2 answers:

5 0

I believe when you mix two things up together. It is called mixture. And they make one Compound. One example could be water, sugar, and flavorings making one compound which is soda. Hope it helps.

Lina20 [59]2 years ago

3 0

Sand and water, sand and iron filings, water and oil

You might be interested in

Place where food is stored and partially digested

Ostrovityanka [42]

Answer: Chyme

Explain: Chyme passes from the stomach to the small intestine. Further protein digestion takes place in the small intestine.

3 0

2 years ago

The sun’s core is made of solid rock. True or false

KengaRu [80]

Answer:

False

Explanation:

Heres what its made of:

The core is made of hot, dense plasma (ions and electrons), at a pressure estimated at 265 billion bar (3.84 trillion psi or 26.5 petapascals (PPa)) at the center. Due to fusion, the composition of the solar plasma drops from 68–70% hydrogen by mass at the outer core, to 34% hydrogen at the core/Sun center.

8 0

2 years ago

2) A common "rule of thumb" -- for many reactions around room temperature is that the

babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$ and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}}$

$\Rightarrow \frac{K_1}{K_2}= e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}= 3.479 \times 10^{-6}$

$\Rightarrow \frac{K_2}{K_1}= e^{3.479 \times 10^{-6}}$

$\Rightarrow \frac{K_2}{K_1}= 1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

8 0

2 years ago

PLEASE HELP ASAP !!

Ivenika [448]

Answer:

2SO₂ + O₂ + 2H₂O -----------> 2H₂SO₄

Theoretical yield of H₂SO₄ = 213 g

percent yield of H₂SO₄ = 94 %

Explanation:

Data Given:

volume of SO₂ = 48.6 L

mass of H₂SO₄ = 200 g

balance equation = ?

theoretical yield = ?

percent yield = ?

Solution:

Part 1:

first we have to write a balance equation for the reaction

SO₂ gas react with water (H₂O) and excess oxygen

The balanced equation is as under

2SO₂ + O₂ + 2H₂O -----------> 2H₂SO₄

Part 2:

Now we have to find theoretical yield

First look at the balance reaction

2SO₂ + O₂ + 2H₂O -----------> 2H₂SO₄

2 mol 2 mol

2 moles of SO₂ give gives 2 moles of H₂SO₄

Now calculate volume of 2 moles of SO₂ and mass of 2 moles of H₂SO₄

volume of 2 moles of SO₂

Formula used

volume of gas = no. of moles x molar volume . . . . . . (1)

molar volume of SO₂= 22.4 L/mol

Put values in above formula (1)

volume of gas = 2 mol x 22.4 L/mol

volume of gas = 44.8 L

volume of 2 mole of SO₂ = 44.8 L

Now,

Find mass of 2 mole H₂SO₄

Formula Used

mass in grams = no. of moles x molar mass . . . . . . . (2)

molar mass of H₂SO₄ = 2 (1) + 32 + 4(16)

molar mass of H₂SO₄ = 98 g/mol

put values in equation 2

mass in grams = 2 mol x 98 g/mol

mass in grams = 196 g

mass of 2 mole of H₂SO₄ = 196 g

** So,

Now we come to know that

44.8 L of SO₂ gives 196 g of H₂SO₄ then how many grams of the H₂SO₄ will be produced by 48.6 L of SO₂

Apply unity Formula

44.8 L of SO₂ ≅ 196 g of H₂SO₄

48.6 L of SO₂ ≅ X g of H₂SO₄

Do cross multiplication

g of H₂SO₄ = 196 g x 48.6 L / 44.8 L

g of H₂SO₄ = 213 g

So that is why the theoretical yield of H₂SO₄ is 213 g

Theoretical yield of H₂SO₄ = 213 g

Part 3

Calculate Percent Yield:

Formula used for this purpose:

percent yield = actual yield /theoretical yield x 100 %

Put value in the above formula

percent yield = 200 g/ 213 g x 100 %

percent yield = 94 %

So percent yield of H₂SO₄ = 94 %

8 0

2 years ago

Which of the following are correct for zero-order reactions?

GREYUIT [131]

Answer:

The answer is "Choice A and Choice B"

Explanation:

The Zero-Order reactions are usually found if a substrate, like a surface or even a catalyst, is penetrated also by reactants. Its success rate doesn't depend mostly on the amounts of the various reaction in this reaction.

Let the Rate = k

As $\frac{dx}{dt} \ rate\ \ K_0$ doesn't depend on reaction rate, a higher reaction rate does not intensify the reaction.

By the rate $k_0 =\frac{dx}{dt},$ the created based and the reaction rate is about the same.

5 0

2 years ago