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murzikaleks [220]
3 years ago
10

Convert 5.4 atm to torr

Chemistry
1 answer:
devlian [24]3 years ago
4 0
1 atm = 760 torr

5.4 atm * (760 torr/1 atm) = 4104 torr

Since 5.4 has 2 significant digits, the final answer must also have 2 significant digits.

The final answer is 4100 torr.
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What is the pH of a buffer solution that is 0.211 M in hypochlorous acid (HClO) and 0.111 M in sodium hypochlorite? The Ka of hy
arsen [322]

Answer:

c. 7.14

Explanation:

The buffer solution is formed by a weak acid ( hypochlorous acid, HClO) and its conjugate base (hypochlorite ClO⁻, coming from sodium hypochlorite NaClO). We can calculate the pH using the Henderson-Hasselbach equation.

pH = pKa + log [base]/[acid]

pH = -log 3.8 × 10⁻⁸ + log 0.111/0.211

pH = 7.14

7 0
3 years ago
If a gas is initially at a pressure of nine ATM and a volume at 21 L at a temperature of 253K and the pressure is raise to 15 AT
Tom [10]

Answer:

15.0L

Explanation:

p/v = constan

(9*21)/253 =(15v)/ 302

v = (9*21*302)/(15*253)

v=15.0

3 0
3 years ago
43 moles feci3 are dissolved in 0.64l of solution what is the molarity of the solution
nasty-shy [4]
Molarity=moles/liter
molarity=43/0.64
molarity=67.19moles/litre
4 0
3 years ago
In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char
lisabon 2012 [21]

Answer : The partial pressure of N_2,H_2\text{ and }NH_3 at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution :  Given,

Initial pressure of N_2 = 1.42 bar

Initial pressure of H_2 = 2.87 bar

K_p = 0.036

The given equilibrium reaction is,

                              N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)

Initially                   1.42      2.87             0

At equilibrium    (1.42-x)  (2.87-3x)     2x

The expression of K_p will be,

K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}

Now put all the values of partial pressure, we get

0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}

By solving the term x, we get

x=0.287\text{ and }3.889

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of NH_3 at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of N_2 at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of H_2 at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of N_2 + Partial pressure of H_2 + Partial pressure of NH_3

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0
3 years ago
Is a flashlight turned on kinetic or potential energy ?
kompoz [17]
Answer: potential energy













8 0
3 years ago
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