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2 years ago

7

A student stores 20.0 moles of gas particles in a flexible container. The student then uses a pump to remove 10.0 moles of gas p

articles. The temperature and pressure remain constant. Which of the following is a true statement?

2 answers:

pav-90 [236]2 years ago

5 0

The volume of the container will decrease.

Sergeeva-Olga [200]2 years ago

3 0

The volume of the container will decrease. I believe this answer is correct because for the temperature and pressure to remain the same but there to be less particles the container must shrink to suit.

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What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

Mashutka [201]

Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

$P_H =- log[H^{+} ]$

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

$[H^{+} ] =0.0014$

$c=0.0014\times \frac{1}{10}$

$c=1.4\times 10^{-4}$

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2 years ago

HIPVs can cause what two more severe illnesses?

Nat2105 [25]

One of them are Cancer

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2 years ago

Calculate the number of grams of solute in 500.0 mL of 0.189 M KOH.

KIM [24]

Answer : The number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams

Solution : Given,

Volume of solution = 500 ml

Molarity of KOH solution = 0.189 M

Molar mass of KOH = 56 g/mole

Formula used :

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

Now put all the given values in this formula, we get the mass of solute KOH.

$0.189M=\frac{\text{Mass of KOH}\times 1000}{(56g/mole)\times (500ml)}$

$\text{Mass of KOH}=5.292g$

Therefore, the number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams

7 0

2 years ago

Read 2 more answers

Suppose a soap manufacturer starts with a triglyceride that has the fatty acid chains arachidic acid, palmitic acid and palmitic

DIA [1.3K]

Answer:

Sodium arachidate; Sodium palmitate and Sodium palmitate

Explanation:

Triglycerides are esters of fatty acids with glycerol. In triglycerides, three fatty acid molecules are linked by ester bonds to each of the three carbon atoms in a glycerol molecule. The fatty acids may be same or different fatty acid molecules. Hydrolysis of triglycerides yields the three fatty acid molecules and glycerol.

Saponification is the process by which a base is used to catalyst the hydrolysis of the ester bonds in glycerides. The products of this base-catalyzed hydrolysis of triglycerides are the metallic salts of the three fatty acids and glycerol. The salts of the fatty acids are known as soaps.

For a triglyceride that has the fatty acid chains arachidic acid, palmitic acid and palmitic acid attached to the three backbone carbons glycerol, the saponification of the triglyceride with NaOH will yield the sodium salts or soaps of the three fatty acids as well as glycerol.

Arachidic acid will react with NaOH to yield sodium arachidate.

The two palmitic acid molecules will each react with NaOH to yield sodium palmitate.

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1 year ago

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Flauer [41]

Explanation:

Dichloromethane is flammable - FALSE

Methanol is flammable. - TRUE

Concentrated sulfuric acid is corrosive. - TRUE

10% sodium carbonate solution must be used in the fume hood. - FALSE

Benzoyl chloride is a lachrymator. - TRUE

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2 years ago