If you are correct you get a brainless

Aluminum is more reactive than iron, yet it is used today for a variety of applications in which iron would corrode (cans, rain

marin [14]

Aluminum is more reactive than iron

but it forms Al-oxide, that form a thin layer on the surface of Al, and protect Al from reaction with water.

Answer: Very unreactive aluminum oxide forms a thin layer on aluminum.

6 0

2 years ago

Glucose is stirred into water to form a solution. In this case, what are the glucose and water?

Alex_Xolod [135]

The correct option is C.

When glucose is stirred into water to form a solution, then glucose is the SOLUTE and water is the SOLVENT. A solute refers to a substance that is dissolved in a solvent. For instance, salt is the solute that is dissolved in salt solution. A solvent on the other hand refers to the liquid into which the solute was dissolved. For instance, in the case of salt solution, water is the solvent, because it is used to dissolve the salt.

5 0

2 years ago

The [OH-] of an aqueous solution is 3.5 x 10-5. What is the [H+]?

Sergeeva-Olga [200]

Answer: The concentration of hydrogen ion in the given solution is $2.81 \times 10^{-10}$ .

Explanation:

pOH of a solution is the negative logarithm of concentration of hydroxide ions.

Hence, if $[OH^{-}]$ is $3.5 \times 10^{-5}$ then its pOH value is calculated as follows.

$pOH = -log [OH^{-}]\\= -log (3.5 \times 10^{-5})\\= 4.45$

Now, the relation between pH and pOH is as follows.

pH + pOH = 14

pH + 4.45 = 14

pH = 14 - 4.45

= 9.55

And, pH is the negative logarithm of concentration of hydrogen ions. Hence, $[H^{+}]$ is calculated as follows.

$pH = -log [H^{+}]\\9.55 = -log [H^{+}]\\H^{+} = 2.81 \times 10^{-10}$

Thus, we can conclude that the concentration of hydrogen ion in the given solution is $2.81 \times 10^{-10}$ .

4 0

2 years ago

A liquid mixture contains 1 kg of water, 1.9 lb of ethanol, and 4.6 lb of methyl acetate. What is the weight fraction of ethanol

slamgirl [31]

<u>Answer:</u> The weight fraction of ethanol in the mixture is

<u>Explanation:</u>

We are given:

Mass of water = 1 kg = 2.205 lb (Conversion factor: 1 kg = 2.205 lb)

Mass of ethanol = 1.9 lb

Mass of ethyl acetate = 4.6 lb

Mass of mixture = [2.205 + 1.9 + 4.6] = 8.705 lb

To calculate the percentage composition of ethanol in mixture, we use the equation:

$\%\text{ composition of ethanol}=\frac{\text{Mass of ethanol}}{\text{Mass of mixture}}\times 100$

Mass of mixture = 8.705 lb

Mass of ethanol = 1.9 lb

Putting values in above equation, we get:

$\%\text{ composition of ethanol}=\frac{1.9lb}{8.705lb}\times 100=21.8\%$

Hence, the weight fraction of ethanol in the mixture is 21.8 %

4 0

2 years ago

If the partial molar volumes of the two components of a binary solution are VA 30 m2/mol and Vg 20 m3/mol and the mole fraction

gregori [183]

Answer : The molar volume of the solution is, $24m^3/mole$

Explanation : Given,

Partial molar volumes of component A = $30m^3/mole$

Partial molar volumes of component B = $20m^3/mole$

Mole fraction of component A = 0.4

First we have to calculate the mole fraction of component B.

As we know that,

$\text{Mole fraction of component A}+\text{Mole fraction of component B}=1$

$\text{Mole fraction of component B}=1-0.4=0.6$

Now we have to calculate the molar volume of the solution.

Expression used :

$V_s=X_A\times \bar V_A+X_B\times \bar V_B$

where,

$\bar V_A$ = partial molar volumes of component A

$\bar V_B$ = partial molar volumes of component B

$V_s$ = molar volume of the solution

$X_A$ = mole fraction of of component A

$X_B$ = mole fraction of of component B

Now put all the give values in the above expression, we get:

$V_s=0.4\times 30m^3/mole+0.6\times 20m^3/mole$

$V_s=24m^3/mole$

Therefore, the molar volume of the solution is, $24m^3/mole$

3 0

2 years ago