What is not a resource of the ocean?

If you needed to make 100 mL of a 0.2 M fruit drink solution from the 1.0 M fruit drink solution, how would you do it? (Hint: Us

Studentka2010 [4]

Answer: We take 20 ml of 1.0 M fruit drink and add 80 ml of water to get 100 ml of 0.2 M solution.

Explanation:

According to the neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = 0.2 M

$V_1$ = volume of stock solution = 100 ml

$M_2$ = molarity of resulting solution = 1.0 M

$V_2$ = volume of resulting solution = ?

Now put all the given values in the above law, we get the volume of resulting solution.

$(0.2M)\times 100=(1.0M)\times (V_2)$

$V_2=20ml$

Therefore, the volume of 1.0 M required is 20 ml.

We take 20 ml of 1.0 M fruit drink and add 80 ml of water to get 100 ml of 0.2 M solution.

7 0

4 years ago

Baking soda, or sodium bicarbonate, is used as a leavening or raising agent in some types of bread. The sodium bicarbonate react

erma4kov [3.2K]

The rate of reaction refers to how fast or slow a reaction is. The rate of reaction can be increased by;

i) Using finely divided sodium carbonate

ii) Increasing the temperature

<h3>Rate of reaction</h3>

Thee term rate of reaction refers to the pace of a chemical reaction. A reaction could be fast or slow.

To increase the rate of this reaction;

i) The sodium bicarbonate could be used in the finely divided form.

ii) The temperature could be raised.

Learn more about rate of reaction: brainly.com/question/13440548

3 0

3 years ago

You calculate the Wilson equation parameters for the ethanol (1) 1 1-propanol (2) system at 258C and find they are L12 5 0.7 and

Gala2k [10]

Here is the correct question.

You calculate the Wilson equation parameters for the ethanol (1) +1 - propanol (2) system at 25° C and find they are ∧₁₂ = 0.7 and ∧₂₁ = 1.1 . Estimate the value of parameters at 50° C

Answer:

the values of the parameters at 50° C are 0.766 and 1.047

Explanation:

From "critical point , enthalpy of phase change and liquid molar volume " the liquid molar volume (v) of ethanol and 1 - propanol is represented as follows:

Compound Liquid molar volume (cm³/mol)

Ethanol (1) 58.68

1 - propanol (2) 75.14

To calculate the temperature dependent parameters of the Wilson equation ∧₁₂.

∧₁₂ = $\frac{V_2}{V_1} \ exp \ (\frac{-a_{12}/R}{T} )$ ------------ equation (1)

where:

$a_{12}/R$ = Wilson parameter = ???

$V_2$ = liquid molar volume of component 2 = 75.14 cm³/mol

$V_1$ = liquid molar volume of component 1 = 58.68 cm³/mol

T = temperature = 25° C = ( 25 + 273.15) K = 298.15 K

Replacing our values in the above equation ; we have:

$0.7 = \frac{75.14 \ cm^3/mol}{58.68 \ cm^3/mol} \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$0.7 = 1.281 \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$In (0.547) = \ (\frac{-a_{12}/R}{298.15 \ K} )$

$-a_{12}/R= 0.60 * 298.15 \ K$

$-a_{12}/R= - 178.89 \ K$

$a_{12}/R= 178.89 \ K$

To calculate the temperature dependent parameters of the Wilson equation ∧₂₁

∧₂₁ = $\frac{V_1}{V_2} \ exp \ (\frac{-a_{12}/R}{T} )$ ---------- equation (2)

$1.1 = \frac{58.68 \ cm^3/mol}{75.14 \ cm^3/mol} \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$1.1 = 0.7809 \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$\frac{1.1}{0.7809}= exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$1n ( 1.4086)= (\frac{-a_{12}/R}{298.15 \ K} )$

$-a_{12}/R = 0.3426 * 298.15 \ K$

$-a_{12}/R =102.15 \ K$

$a_{12}/R = -102.15 \ K$

From equation (1) ; let replace 178.98 K for $a_{12}/R$

$V_2 =$ 75.14 cm³/mol

$V_1$ = 58.68 cm³/mol

T = 50° C = ( 50 + 273.15 ) K = 348.15 K

So;

∧₁₂ = $\frac{75.14 \ cm^3/mol}{58.68 \ cm^3/mol} \ exp \ (\frac{- 178,.89 \ K}{348.15 \ K} )$

∧₁₂ = 1.281 exp(-0.5138)

∧₁₂ = 1.281 × 0.5982

∧₁₂ =0.766

From equation 2; let replace 102.15 K for $a_{12}/R$

$V_2 =$ 75.14 cm³/mol

$V_1$ = 58.68 cm³/mol

T = 50° C = ( 50 + 273.15 ) K = 348.15 K

So;

∧₂₁ = $\frac{58.68 \ cm^3/mol}{75.14 \ cm^3/mol} \ exp \ (\frac{-(-102.15)\ K}{298.15 \ K} )$

∧₂₁ = 0.7809 exp (0.2934)

∧₂₁ = 0.7809 × 1.3410

∧₂₁ = 1.047

Thus, the values of the parameters at 50° C are 0.766 and 1.047

8 0

3 years ago

Activation energy of a reaction: is an alternative term for the free energy of activation. is decreased in the presence of an en

Fynjy0 [20]

Explanation:

Activation energy, in chemistry, is the minimum amount of energy that is required to activate atoms or molecules to a condition in which they can undergo chemical transformation or physical transport. Basically, activation energy is that push that atoms need in order to undergo reactions.

All chemical reactions, including exothermic reactions, need activation energy to get started. Activation energy is needed so reactants can move together, overcome forces of repulsion, and start breaking bonds.

A catalyst lowers the activation energy of a reaction, so that a chemical reaction can take place. Increasing the temperature of a reaction has the effect of increasing the number of reactant particles that have more energy than the activation energy. Enzymes are a form of catalysts that speed up chemical reactions by lowering the activation energy.

The correct options are therefore;

- is decreased in the presence of an enzyme specific for that reaction.

- is the amount of free energy required to bring the reactants to the transition state.

- is an alternative term for the free energy of activation.

5 0

4 years ago

In a laboratory experiment, after purification, 4.567 g of nickel metal was obtained by a student. The student then calculated,

DaniilM [7]

Answer:

The yield of the purification is 79.7%

Explanation:

Hi there!

5.731 g nickel metal could be produced by the student if the yield of the purification would be 100%. Instead, 4.567 g was obtained. Then, the yield of the purification can be calculated as follows:

4.567 g Ni · (100% / 5.731 g) = 79.7%

The yield of the purification is 79.7%

Have a nice day!

5 0

4 years ago

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