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zzz [600]
3 years ago
13

A tank of methane gas contains 2.8 m³ of the gas at 20°C. The tank has a pressure-release valve that releases gas into a seconda

ry tank that will hold 0.2m³ of gas if the pressure rises.
At what Celsius temperature will the methane fill both tanks?
Chemistry
1 answer:
iren2701 [21]3 years ago
5 0

Answer:

The answer to your question is T2 = 20.93°K

Explanation:

Data

Volume 1 = V1 = 2.8 m³

Temperature 1 = T1 = 20°C

Volume 2 = V2 = 0.2 m³

Temperature 2 = T2 = ?

Process

To solve this problem use Charles' law

               \frac{V1}{T1} = \frac{V2}{T2}

Solve for T2

               T2 = \frac{T1V2}{V1}

1.- Convert temperature to °K

T1 = 20 + 273 = 293°K

2.- Substitute values

               T2 = \frac{(293)(0.2)}{2.8}

3.- Simplify

               T2 = \frac{58.6}{2.8}

4.- Result

               T2 = 20.92°K

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Pleasseee help!! Got STUCK FOR A WHILE!! How does this equation help you to understand acids and bases: H+(aq) + OH-(aq)H2O(aq)?
Gekata [30.6K]

Answer:

Odd answer choices, but I would put "It shows you that water is a reversible reaction, which contains hydrogen and hydroxide reactions."

Explanation:

The first choice is weird, although water can be written as HOH that doesn't help you understand acids and bases.

The third choice doesn't do that either, and there are more ways for water to be formed than that.

The second choice helps you understand acids and bases, specifically, in water's self-ionization process. It also shows you that it is neutral because it is formed from Hydrogen, which is usually present in acids, and a Hydroxide group, which is usually present in bases.

7 0
3 years ago
In preparation for a blizzard, a city dispatches a crew to spread salt on the street surface of the bridges around town. The sal
Ilia_Sergeevich [38]
0 degrees to -6 or -16 degrees Celsius

3 0
4 years ago
A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base
tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ =  0.00375  mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

HN_3 + OH- ---> N^-_{3} + H_2O

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction =  0.025 +  0.0133 = 0.0383  L

Concentration of HN_3 = \dfrac{0.001755}{0.0383} = 0.0458 M

Concentration of N^{-}_3 = \dfrac{ 0.001995 }{0.0383} = 0.0521 M

GIven that :

Ka = 1.9 x 10^{-5}

Thus; it's pKa = 4.72

pH =4.72 +  log(\dfrac{ \ 0.0521}{0.0458})

pH =4.72 + log(1.1376)

pH =4.72 + 0.05598

pH =4.77598

pH ≅ 4.80

3 0
3 years ago
How many grams of n2f4 can be produced from 225g f2?
3241004551 [841]
The answer is 615.91 grams of <span>n2f4

Solution:
225g F2 x [(1molF2)/(38gramsF2)] x [</span>(1molF2)/(1molN2F4)] x [(104.02 grams N2F4)/(1molN2F4)]
=615.91 grams
8 0
4 years ago
If the ph of hc3h5o2 is 4.2 and the ka 1.34x10^-5, what is the equilibrium concentration
n200080 [17]
<span>CH</span>₃<span>CH</span>₂<span>COOH + H</span>₂<span>O </span>↔ <span> CH</span>₃<span>CH</span>₂<span>COO</span>⁻<span> + H</span>₃<span>O</span>⁺<span> 
</span>
pH = 0.5 pKa + 0.5 pCa
0.5 pCa = pH - 0.5 pKa
              = 4.2 - (0.5 * (-log 1.34 x 10⁻⁵)) = 1.76
pCa = 3.53
Ca = antilog - 3.52 = 3 x 10⁻⁴ 
where Ca is the acid concentration 
5 0
3 years ago
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