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EastWind [94]
3 years ago
8

A sample of a gas originally at 29 °c and 1.25 atm pressure in a 3.0 l container is allowed to contract until the volume is 2.2

l and the temperature is 11 °c. the final pressure of the gas is ________ atm.
Chemistry
1 answer:
Anastasy [175]3 years ago
5 0
<span>Answer: 1.6 atm. pV/T = constant. (1.25 * 3.0)/(273+29) <-- work in Kelvin This gives you the constant. Then you want p So p = (T*constant)/V = (284*0.0124)/2.2 I get = 1.6atm</span>
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Which of the following Kb values represents the strongest base?
alexandr402 [8]

Answer:

D. Kb = 1.8 × 10⁻⁵  

Explanation:

The strongest base has the largest Kb value.

The largest Kb value has the smallest negative exponent.

So, the strongest base has Kb = 1.8 × 10⁻⁵.

4 0
3 years ago
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Using the table above, find the average atomic mass of the element. Then, using the periodic table, determine which element is s
lara [203]

Answer:

Chlorine

Explanation:

  • The abundance of chlorine-35 is 75% and the abundance of chlorine-37 is 25%.
  • Chlorine for example has two naturally occurring isotopes: Chlorine-35 and chlorine-37.
  • Chlorine's atomic mass is 35.453 u.
7 0
3 years ago
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An ideal gas is allowed to expand from 5.40 L to 35.1 L at constant temperature. By what factor does the volume increase?
horrorfan [7]

Answer:

a) 6.5

b) pressure decreases

c)22 atm

Explanation:

Since the initial volume V1 is 5.40 L and the final volume V2 is 35.1 L, the ratio of V2:V1= 35.1/5.40= 6.5 hence the volume increases by a factor of 6.5.

When the volume increases, the pressure decreases accordingly in accordance with Boyle's law. Boyle's law states that the volume of a given mass of gas is inversely proportional to its pressure at constant temperature.

c) From Boyle's law

Initial volume V1= 5.40 L

Final volume V2= 35.1 L

Initial pressure P1= 143 ATM

Final pressure P2 = the unknown

P1V1= P2V2

P2= P1V1/V2

P2 = 143 × 5.40/ 35.1

P2= 22 atm

6 0
3 years ago
Which combination may be used to prepare a buffer having a ph of 8. 8?
7nadin3 [17]

The combination used in the preparation of a buffer with a pH around 8.8 accounts for ka = 7 * 10 -3 for h 3po4

The ph of the buffer can be shown as:

pH = pKa + log [Salt] /[ Acid ]

[Salt] /[ Acid ] = x

For h3po4 with ka= 7 × 10–3

8.8 = - log (7 × 10^–3) + log x

8.8 = 2.21 + log x

Thus, the value of log x is coming positive and therefore can be used for preparing buffer.

For h2po4- with ka= 8 × 10–8

8.8 = - log (8 × 10^–8) + log x

8.8 = 7.14 + log x

Thus, the value of log x is coming positive and therefore can be used for preparing buffer.

For hpo42– with ka= 5 × 10–13

8.8 = - log (5 × 10–13) + log x

8.8 = 12.31 + log x

Thus, the value of log x is coming negative and therefore can not be used for preparing buffer.

Hence, the correct answer is option A

Learn more about buffering systems here,

brainly.com/question/16556401

# SPJ4

4 0
2 years ago
the solubility product Ag3PO4 is: Ksp = 2.8 x 10^-18. What is the solubility of Ag3PO4 in water, in moles per liter?
guapka [62]

Answer : The solubility of Ag_3PO_4 in water is, 1.8\times 10^{-5}mol/L

Explanation :

The solubility equilibrium reaction will be:

Ag_3PO_4\rightleftharpoons 3Ag^++PO_4^{3-}

Let the molar solubility be 's'.

The expression for solubility constant for this reaction will be,

K_{sp}=[Ag^{+}]^3[PO_4^{3-}]

K_{sp}=(3s)^3\times (s)

K_{sp}=27s^4

Given:

K_{sp} = 2.8\times 10^{-18}

Now put all the given values in the above expression, we get:

K_{sp}=27s^4

2.8\times 10^{-18}=27s^4

s=1.8\times 10^{-5}M=1.8\times 10^{-5}mol/L

Therefore, the solubility of Ag_3PO_4 in water is, 1.8\times 10^{-5}mol/L

3 0
3 years ago
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