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2 years ago

How is the mass in grams of the element converted to amount in atoms

1 answer:

8 0

A substance's molar mass is calculated by multiplying its relative atomic mass by the molar mass constant (1 g/mol). ... By multiplying a given mass by the molar mass, the amount of moles of the substance can be calculated.

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How many moles of HCl are required to neutralize 20.0 mL of 1.5 M KOH?

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2 years ago

when you receive fluid in the hospital, it is often 0.9% saline solution. what would happen if the nurse used distilled water in

jasenka [17]

Answer:

The distilled water solution would be hypotonic to your blood cells, so they would gain water, swell, and possibly burst.

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2 years ago

Acetic acid (HC2H3O2, Ka=1.8x10^-5) is a weak acid. Calculate the pH of an aqueous solution of .25M acetic acid.

tensa zangetsu [6.8K]

Answer: The pH of an aqueous solution of .25M acetic acid is 2.7

Explanation:

$HC_2H_3O_2\rightarrow H^+C_2H_3O_2^-$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 0.25 M and $\alpha$ = ?

$K_a=1.8\times 10^{-5}$

Putting in the values we get:

$1.8\times 10^{-5}=\frac{(0.25\times \alpha)^2}{(0.25-0.25\times \alpha)}$

$(\alpha)=0.0084$

$[H^+]=c\times \alpha$

$[H^+]=0.25\times 0.0084=0.0021$

Also $pH=-log[H^+]$

$pH=-log[0.0021]=2.7$

Thus pH is 2.7

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2 years ago

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2 years ago

Positive Deviation from Raoult's Law occurs when the vapour pressure of component is greater than what is expected in Raoult's L

8090 [49]

Answer:

All of the above.

Explanation:

In positive deviation from Raoult's Law occur when the vapour pressure of components is greater than what is expected value in Raoult's law.

When a solution is non ideal then it shows positive or negative deviation.

Let two solutions A and B to form non- ideal solutions.let the vapour pressure of component A is $P_A$ and vapour pressure of component B is $P_B$ .

$P^0_A$ = Vapour pressure of component A in pure form

$P^0_B$ = Vapour pressure of component B in pure form

$x_A$ =Mole fraction of component A

$x_B=$ =Mole fraction of component B

The interaction between A- B is less than the interaction A- A and B-B interaction.Therefore, the escaping tendency of liquid molecules in mixture is greater than the escaping tendency in pure form.Hence, the vapour pressure of a mixture is greater than the initial value of vapour pressure.

$P_A >P^0_A\cdot x_A$ , $P_B>P^0_B\cdot x_B$

Therefore, $P_A+P_B >P^0_A\cdot x_A+P^0_B \cdot x_B$

Therefore, the enthalpy of mixing is greater than zero and change in volume is greater than zero.

Hence, option a,b,c and d are true.

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2 years ago