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3 years ago

Explain why the pressure exerted by a gas does not depend on the type of the gas.

1 answer:

8 0

Since no matter what type of gas there is there is still a pressure exerted by it. Since the Kinetic theory of gases says " the pressure is proportional to the number of particles in the gas".
Basically pressure exerted by the gas depends rather on the temperature. the temperature increases, and the volume still says the same though.

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State one advantage of building a space station in orbit instead of sending it all up in one piece.

Anit [1.1K]

Building a space station in orbit is a safety measure. If a rocket fails to bring up the entire space station in one piece it would be extremely expensive to build a new one. Bringing it up in multiple places ensures less money would be lost if the rocket fails.

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3 years ago

The word root for a group of microorganisms living together is a. steriliz. b. prodrom. c. coloniza. d. nosocomi.

UkoKoshka [18]

The answer is C) Coloniza.

7 0

3 years ago

(a) Kw = 1.139 × 10⁻¹⁵ at 0°C and 5.474 × 10⁻¹⁴, find [H₃O⁺] and pH of water at 0°C and 50°C.

kondor19780726 [428]

The value of [H₃O⁺] and pH of water at 0°C and 50°C.

0°C value of [H₃O⁺] = 3.375 x 10⁻⁸

50°C value of [H₃O⁺] = 2.340 x 10⁻⁷

pH of water at 0°C = pH = 7.4717

pH of water at 50°C = pH = 6.6308

<h3>pH of water at different level:</h3>

Water with a pH less than 7 is considered acidic, while water with a pH greater than 7 is considered basic. The normal pH range for surface water systems is 6.5 to 8.5, and for groundwater systems is 6 to 8.5. Alkalinity is a measure of a water's ability to withstand a pH change that would cause it to become more acidic.

<h3>According to the given information:</h3>

0°C = Kw = 1.139 × 10⁻¹⁵

50°C = 5.474 × 10⁻¹⁴

Solving at 0°C for [H₃O⁺] and pH. water is neutral therefore its[H₃O⁺] [OH⁻]

are equal [H₃O⁺] = [OH⁻].

Kw = [H₃O⁺] [OH⁻]

Kw = [H₃O⁺]²

[H₃O⁺] = √Kw

= √1.139 × 10⁻¹⁵

= 3.375 x 10⁻⁸

pH = -log[H₃O⁺]

= -log 3.375 x 10⁻⁸

= 7.4717

Solving at 50°C for [H₃O⁺] and pH. water is neutral therefore its[H₃O⁺] [OH⁻]

are equal [H₃O⁺] = [OH⁻].

Kw = [H₃O⁺] [OH⁻]

Kw = [H₃O⁺]²

[H₃O⁺] = √Kw

= √ 5.474 × 10⁻¹⁴

= 2.340 x 10⁻⁷ M

pH = -log[H₃O⁺]

= -log2.340 x 10⁻⁷

pH = 6.6308

The value of [H₃O⁺] and pH of water at 0°C and 50°C.

0°C value of [H₃O⁺] = 3.375 x 10⁻⁸

50°C value of [H₃O⁺] = 2.340 x 10⁻⁷

pH of water at 0°C = pH = 7.4717

pH of water at 50°C = pH = 6.6308

To know more about pH of water visit:

brainly.com/question/13822050

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I understand that the question you are looking for is:

Kw = 1.139 × 10⁻¹⁵ at 0°C and 5.474 × 10⁻¹⁴, find [H₃O⁺] and pH of water at 0°C and 50°C. find the [H₃O⁺] and pH of water at 0°C and 50°C.

3 0

2 years ago

Which response includes all the following processes that are accompanied by an increase in entropy? 1) 2SO 2(g) + O 2(g) → SO 3(

Mashutka [201]

Answer: Reaction (1) , (3) and (4) are accompanied by an increase in entropy.

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

(1) $2SO_2(g)+O_2(g)\rightarrow SO_3(g)$

3 moles of reactant are changing to 1 mole of product , thus the randomness is increasing. Thus the entropy also increases.

2) $H_2O(l)\rightarrow H_2O(s)$

1 mole of Liquid reactant is changing to 1 mole of solid product , thus the randomness is decreasing. Thus the entropy also decreases.

3) $Br_2(l)\rightarrow Br_2(g)$

1 mole of Liquid reactant is changing to 1 mole of gaseous product , thus the randomness is increasing. Thus the entropy also increases.

4) $H_2O_2(l)\rightarrow H_2O(l)+\frac{1}{2}O_2(g)$

1 mole of Liquid reactant is changing to half mole of gaseous product and 1 mole of liquid product, thus the randomness is increasing. Thus the entropy also increases.

4 0

3 years ago

Please help me thank you.

Semmy [17]

I will show you with detailed work for NaCl, but follow the same procedure for the rest of the compounds.

Molar Mass - Find the molar mass of the Na and the Cl and add them together
Na - 23
Cl - 35.5
Add those numbers together 23 + 35.5 = 58.5 g/mol

Moles in 1 tsp:
The mass measured in 1tsp of NaCl was 18 g. To calculate the amount of moles you take the mass measured and divide it by the molecular weight.
18/58.5 = 0.3077 mol

Moles of each element:
To find the moles each element in the compound you multiply the moles of 1 tsp by the number of atoms of the element in the compound
Na - 1 in NaCl
Cl - 1 in Na Cl
so take 0.3077 * 1 = 0.3077 moles Na (and Cl in this case)

Atoms of each:
take the number of moles calculated and multiply that by Avogadro's number(6.023x10^23) for the number of molecules
So for both Na and Cl:
0.3077 * 6.023x10^23 = 1.853x10^23 atoms for both Na and Cl

7 0

4 years ago