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amid [387]
1 year ago
12

A 4.27 g sample of a lab solution contains 1.81g of acid. What is the concentration of the solution as a mass percentage?

Chemistry
1 answer:
vaieri [72.5K]1 year ago
5 0

Answer:

42.38875878%

Explanation:

i divided 4.27 g from 1.81g using a percentage calculator but im not sure if its correct

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Draw arrows for all of the attractive and repulsive forces between the negatively charged electrons and positively charged proto
Sophie [7]

Answer:

Attraction

(e-) ---> <--- (H+)

Repulsion:

<---(e-) (e-)-->

Neutral:

(e-)       (Helium)

Explanation:

Accordingly to coulomb's law:

In the attraction, the hydrogen without an electron has a positive charge and needs to be fulfilled with a negative charge found in an eletron.

In the repulsion, both electrons has the same charge and repulse each other.

In the neutral case, the Helium is highly stable therefore the electron is not attracted.

4 0
2 years ago
The branch of chemistry that looks at the release of electrical energy is called
sergij07 [2.7K]
B is the correct answer tbh
3 0
3 years ago
what do all word equations start with? what comes next in the word equation? how is the word equation concluded ?
Sladkaya [172]
They start with  the numbers u need to know in order to slove the problem and there has to be a story behind it 
7 0
3 years ago
You wish to make a 0.299 M hydroiodic acid solution from a stock solution of 6.00 M hydroiodic acid. How much concentrated acid
Art [367]

Answer:

V_1=2.49mL

Explanation:

Hello,

In this case, considering that the moles of hydrioiodic acid remain unchanged during the dilution process:

n_{HI}=n_{HI}

One could apply the following equality in terms of molarity:

V_1M_1=V_2M_2

Whereas the subscript 1 accounts for the solution before the dilution and 2 after the dilution, therefore, the required volume of 6.00 M acid is:

V_1=\frac{V_2M_2}{M_1} =\frac{50.0mL*0.299M}{6.00M}=2.49mL

Best regards.

5 0
2 years ago
Find the pHpH of a solution prepared from 1.0 LL of a 0.15 MM solution of Ba(OH)2Ba(OH)2 and excess Zn(OH)2(s)Zn(OH)2(s). The Ks
charle [14.2K]

Answer:

pH  = 13.09

Explanation:

Zn(OH)2 --> Zn+2 + 2OH-   Ksp = 3X10^-15

Zn+2 + 4OH-   --> Zn(OH)4-2   Kf = 2X10^15

K = Ksp X Kf

  = 3*2*10^-15 * 10^15

  = 6

Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M

                Zn(OH)₂ + 2OH⁻(aq)  --> Zn(OH)₄²⁻(aq)

Initial:           0             0.3                      0

Change:                      -2x                     +x

Equilibrium:               0.3 - 2x                 x

K = Zn(OH)₄²⁻/[OH⁻]²

6 = x/(0.3 - 2x)²  

6 = x/(0.3 -2x)(0.3 -2x)

6(0.09 -1.2x + 4x²) = x

0.54 - 7.2x + 24x² = x

24x² - 8.2x + 0.54 = 0

Upon solving as quadratic equation, we obtain;

x = 0.089

Therefore,

Concentration of (OH⁻) = 0.3 - 2x

                                    = 0.3 -(2*0.089)

                                  = 0.122

pOH = -log[OH⁻]

         = -log 0.122

          = 0.91

pH = 14-0.91

     = 13.09

4 0
2 years ago
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