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Helga [31]
2 years ago
4

g Chapter 18 Values The following values will be useful for problems in this chapter. Acid K Substance or Species K HF Ka = 7.2

´ 10-4 NH3 Kb = 1.8 ´ 10-5 HNO2 Ka = 4.5 ´ 10-4 (CH3)3N Kb = 7.4 ´ 10-5 CH3COOH Ka = 1.8 ´ 10-5 [Co(OH2)6]2+ Ka = 5.0 ´ 10-10 HOCl Ka = 3.5 ´ 10-8 [Fe(OH2)6]2+ Ka = 3.0 ´ 10-10 HOBr Ka = 2.5 ´ 10 -9 [Fe(OH2)6]3+ Ka = 4.0 ´ 10-3 HOCN Ka = 3.5 ´ 10-4 [Be(OH2)4]2+ Ka = 1.0 ´ 10-5 HCN Ka = 4.0 ´ 10-10 [Cu(OH2)4]2+ Ka = 1.0 ´ 10-8 H2SO4 Ka1 = very large HBO2 Ka = 6.0 ´ 10-10 Ka2 = 1.2 ´ 10-2 (COOH)2 Ka1 = 5.9 ´ 10-2 H2CO3 Ka1 = 4.2 ´ 10-7 Ka2 = 6.4 ´ 10-5 Ka2 = 4.8 ´ 10-11 CH3NH2 K b = 5.0 ´ 10-4 ***************************************************************************** Calculate the pH of a solution in which [OH-] = 2.50 ´ 10-4M
Chemistry
1 answer:
bekas [8.4K]2 years ago
7 0

Answer:

yes

Explanation:

gergtehwh

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Chlorine is used to disinfect swimming pools. the accepted concentration for this purpose is 1.00 ppm chlorine, or 1.00 g of chl
frez [133]

The accepted concentration of chlorine is 1.00 ppm that is 1 gram of chlorine per million of water.

The volume of water is 2.29\times 10^{4} gal.

Since, 1 gal= 3785.41 mL

Thus, 2.29\times 10^{4} gal=2.29\times 10^{4}\times 3785.41 mL=8.66\times 10^{7}mL

Density of water is 1 g/mL thus, mass of water will be 8.66\times 10^{7}g.

Since, 1 grams of chlorine →10^{6} grams of water.

1 g of water →10^{-6} g of chlorine and,

8.66\times 10^{7}g of water →86.6 g of chlorine

Since, the solution is 9% chlorine by mass, the volume of solution will be:

V=\frac{100}{9}\times 86.6 mL=9.62\times 10^{2} mL

Thus, volume of chlorine solution is 9.62\times 10^{2} mL.

6 0
2 years ago
Which electron has a higher probability of being found at 4 Å of the nucleus, one in a 2s orbital or one in a 2p orbital?
Lena [83]

Answer:

one in a 2s orbital

Explanation:

Because of the peak near the nucleus in the 2s curve there is a higher probability of finding a 2s within 4 Å of the nucleus. In a multi-electron atom an electron in a 2s orbital will have a lower energy than one in a 2p orbital

4 0
2 years ago
Question 17 In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas t
mars1129 [50]

Answer:

Explanation:

N₂       + 3H₂     =     2 NH₃

1 vol                         2 vol

786 liters               1572 liters

786 liters of dinitrogen will result in the production of 1572 liters of ammonia

volume of ammonia V₁ = 1572 liters

temperature T₁ = 222 + 273 = 495 K

pressure = .35 atm

We shall find this volume at NTP

volume V₂ = ?

pressure = 1 atm

temperature T₂ = 273

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

\frac{.35\times 1572}{495} =\frac{1\times V_2}{ 273 }

V_2 =303.44 liter .

mol weight of ammonia = 17

At NTP mass of 22.4 liter of ammonia will have mass of 17 gm

mass of 303.44 liter of ammonia will be equal to (303.44 x 17) / 22.4 gm

= 230.28 gm

=.23 kg / sec .

Rate of production of ammonia = .23 kg /s .

5 0
2 years ago
The density of ice is 0.917 g/cm3. How much volume does 52.3 g of ice occupy? Show your work.
mash [69]

Answer:

The answer is

<h2>57.0 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density}

From the question

mass of ice = 52.3 g

density = 0.917 g/cm³

The volume is

volume =  \frac{52.3}{0.917}  \\  = 57.0338058

We have the final answer as

<h3>57.0 mL</h3>

Hope this helps you

6 0
2 years ago
In a car piston shown above, the pressure of the compressed gas (red) is 5.00 atm. If the area of the piston is 0.0760 m^2. What
Shkiper50 [21]

Answer:

38503.5N

Explanation:

Data obtained from the question include:

P (pressure) = 5.00 atm

Now, we need to convert 5atm to a number in N/m2 in order to obtain the desired result of force in Newton (N). This is illustrated below:

1 atm = 101325N/m2

5 atm = 5 x 101325 = 506625N/m^2

A (area of piston) = 0.0760 m^2

Pressure is force per unit area. Mathematically it is written as

P = F/A

F = P x A

F = 506625 x 0.0760

F = 38503.5N

Therefore, the force exerted on the piston is 38503.5N

6 0
2 years ago
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