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Nuetrik [128]
3 years ago
13

Enter your answer in the provided box.

Chemistry
1 answer:
Lina20 [59]3 years ago
6 0

Answer:

5.06atm

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (Litres)

V2 = final volume (Litres)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question;

P1 = 1.34 atm

P2 = ?

V1 = 5.48 L

V2 = 1.32 L

T1 = 61 °C = 61 + 273 = 334K

T2 = 31 °C = 31 + 273 = 304K

Using P1V1/T1 = P2V2/T2

1.34 × 5.48/334 = P2 × 1.32/304

7.34/334 = 1.32P2/304

Cross multiply

334 × 1.32P2 = 304 × 7.34

440.88P2 = 2231.36

P2 = 2231.36/440.88

P2 = 5.06

The final pressure is 5.06atm

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Answer:

                Option-C (27.36% Na, 1.20% H, 14.30% C, and 57.14% O)

Explanation:

<em>Percent Composition</em> is defined as the <u><em>%age by mass of each element present in a compound</em></u>. Therefore, it is a relative amount of each element present in a compound.

Calculating Percent Composition of NaHCO₃:

1: Calculating Molar Masses of all elements present in NaHCO₃:

              a) Na  =  22.99 g/mol

             b) H  =  1.01 g/mol

              c) C  =  12.01 g/mol

              d) O₃  =  16.0 × 3 =  48 g/mol

2: Calculating Molecular Mass of NaHCO₃:

              Na  =  22.99 g/mol

             H    =  1.01 g/mol

              C    =  12.01 g/mol

              O₃  =  48 g/mol

                       ----------------------------------  

Total                  84.01 g/mol

3: Divide each element's molar mass by molar mass of NaHCO₃ and multiply it by 100:

For Na:

                 =  22.99 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  27.36 %

For H:

                 =  1.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  1.20 %

For C:

                 =  12.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  14.29 % ≈ 14.30 %

For O:

                 =  48.0 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  57.13 % ≈ 57.14 %

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What is the poH of a<br> 2.6 x 10-6 M H+ solution?
andriy [413]

Answer:

Explanation:

pH and pOH.....

The pH is a way of expressing the hydrogen ion concentration.

pH = -log[H+] ............. where [x] means "the concentration of x in moles per liter."

From pH you can compute pOH since at 25C pH + pOH = 14.00 .......... (but only at 25C)

pH = -log(2.6x10^-6) = 5.585 ..... which should be rounded to two significant digits: pH = 5.59

When taking the log of a number, only the digits to the right of the decimal reflect the precision in the original number. Since 2.6x10^-6 has two significant digits, a pH of 5.59 has two significant digits.

pOH + pH = 14.00

pOH = 14.00 - pH = 14.00 - 5.59 = 8.41 ......... at 25C

We can also use the H+ ion concentration to get the hydroxide ion concentration and from that the pOH.

Kw = [H+][OH-] = 1.00x10^-14 .......... at 25C .... like any Kc, the value changes with temperature

[OH-] = Kw / [H+] = 1.00x10^-14 / 2.6x10^-6 = 3.846x10^-9 .... to a couple of guard digits

pOH = -log[OH-] = -log(3.846x10^-9) = 8.415 ...... round to two significant digits: pOH = 8.42 ..... at 25C

=========

Just for grins, you might want to know how Kw changes with temperature, and how [H+] and [OH-] are related at some other temperatures. The pH is the pH of a neutral solution at various temperatures. For instance at 10C a neutral solution has a pH of 7.27. That's not a basic pH. 7.27 is the pH of a neutral solution, but at a different temperature. In a neutral solution at 10C [H+] = [OH-] = 5.41x10^-8M.

pH and Kw for a neutral solution at different temperatures

T .........pH ......... Kw

0......... 7.47....... 0.114 x 10-14

10....... 7.27....... 0.293 x 10-14

20....... 7.08....... 0.681 x 10-14

25....... 7.00....... 1.008 x 10-14

30....... 6.92....... 1.471 x 10-14

40....... 6.77....... 2.916 x 10-14

50....... 6.63....... 5.476 x 10-14

100..... 6.14....... 51.3 x 10-14

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