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ololo11 [35]
3 years ago
14

An ice freezer behind a restaurant has a freon leak, releasing 41.60 g of C2H2F3Cl into the air every week. If the leak is not f

ixed, how many kilograms of fluorine will be released into the air over 6 months? Assume there are 4 weeks in a month. mass of fluorine leaked over 6 months:
Chemistry
1 answer:
zlopas [31]3 years ago
5 0
Weeks = 6 x 4 = 24
Mass leak rate of freon = 41.60 g/week
Mass leak rate of fluorine
Fluorine mass in Freon
= —————————————- X leak rate
M.M. Of Freon

19 x 3
——— X 41.60 = 20.010 gm/week
118.5

Total leaked in 6 months

= 24 x 20.010 = 480.24 gm = 0.480 Kg
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<span>The symbol for hydronium ion concentration is H+. </span><span>There are quite a few relationships between [H+] and [OH−] ions. And because there is a large range of number between 10 to 10</span><span>-15</span><span> M, the pH is used. pH = -log[H+] and pOH = -log[OH−]. In aqueous solutions, </span><span>[H+ ][OH- ] = 10-14. From here we can derive the values of each concentration.</span>
3 0
3 years ago
Which object can electrically polarized
kobusy [5.1K]

Answer:Static electricity works because objects which are otherwise "neutral" (in other words, objects with no net charge) can be polarized. An electric field, like one caused by a nearby charged object, can cause the charges inside of a neutral object — the protons and electrons — to move around a tiny bit.

Explanation:

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3 years ago
You have 4 moles of oxygen gas in a flask. 4 moles of helium gas is added. What happens to the total pressure of the gases in th
ASHA 777 [7]

Answer: The correct option is (c). The total pressure doubles.

Solution:

Initially,  only 4 moles of oxygen gas were present in the flask.

p_{O_2}=Tp_1\times X_{O_2}  (X_{O_2}=\frac{4}{4}) ( according to Dalton's law of partial pressure)

p_{O_2}=Tp_1\times 1=Tp_1....(1)

Tp_1= Total pressure when only oxygen gas was present.

Final total pressure when 4 moles of helium gas were added:

X'_{O_2}=\frac{4}{8}=\farc{1}{2},X_{He}=\frac{4}{8}=\frac{1}{2}

partial pressure of oxygen in the mixture :

Since, the number of moles of oxygen remains the same, the partial pressure of oxygen will also remain the same in the mixture.

p_{O_2}=Tp_2\times X'_{O_2}=Tp_2\times \frac{1}{2}

Tp_2= Total pressure of the mixture.

from (1)

Tp_1=Tp_2\times X'_{O_2}=Tp_2\times \frac{1}{2}

On rearranging, we get:

Tp_2=2\times Tp_1

The new total pressure will be twice of initial total pressure.

7 0
3 years ago
A particular radioactive nuclide has a half-life of 1000 years. What percentage of an initial population of this nuclide has dec
Delicious77 [7]

Answer:

91.16% has decayed & 8.84% remains

Explanation:

A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt

Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹

Time (t) = 1000yrs  

A = fraction of nuclide remaining after 1000yrs

A₀ = original amount of nuclide = 1.00 (= 100%)  

lnA = lnA₀ - kt

lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426

A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years

Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.

3 0
3 years ago
The gas in a cylinder has a volume of 8 liters at a pressure of 101 kPa. The pressure of the gas is increased to 206 kPa. Assumi
Arlecino [84]
The pressure of the gas is increased<span> to 224 </span><span>kPa</span>
3 0
3 years ago
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