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2 years ago

5

The mass of the electron is..

2 answers:

musickatia [10]2 years ago

5 0

The answer to the question is b

Molodets [167]2 years ago

3 0

The correct answer is b, almost the same as a the proton

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A solution is created by dissolving 11.0 grams of ammonium chloride in enough water to make 235 mL of solution. How many moles o

Amanda [17]

Answer:

The correct answer is 0.206 moles

Explanation:

According to the given scenario, the calculation of the number of moles of ammonium chloride is available in the resulting solution is given below:

Given that

Amount of $NH_4Cl$ is 11.0 grams

And, the volume is 235 mL

Now the molar mass of $NH_4Cl$ is 53.49g/mol

So, the number of moles presented is

= 11.0 ÷ 53.49

= 0.206 moles

hence, the number of moles of ammonium chloride are available in the resulting solution is 0.206 moles

7 0

2 years ago

The chemical formula for copper (II) phosphate is Cu3(PO4)2. What is the charge on each copper ion?

mrs_skeptik [129]

3+ because I just know it 8757890

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2 years ago

How are all atoms of silver alike, but different from atoms of other elements?

andriy [413]

B. All atoms of silver have the same atomic number but different numbers of neutrons in the nucleus.

7 0

2 years ago

What would be the pH of a solution of a solution with a hydrogen ion concentration of 0.001 M?

son4ous [18]

0001 M HCl is the same as saying that 1 *10-4 moles of H+ ions have been added to solution. The -log[. 0001] =4, so the pH of the solution =4.

7 0

2 years ago

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 ;

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

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

$[H^{+} ] =0.0014$

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

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

6 0

3 years ago