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strojnjashka [21]
3 years ago
9

What is the approximate pressure of a storage cylinder of recovered R-404A that does not contain any non-condensable impurities

and is stored in a room where the temperature is 80°F?
Chemistry
2 answers:
Aloiza [94]3 years ago
8 0

Answer:

The answer is 173 psig

Explanation:

R404A is a mixture consisting of R-125, R-143A and R-134a. It is the substitute for R-502 for the refrigeration sector in installations for low and medium temperatures. It is characterized by its chemical stability. Its application is the new installations for low and medium temperatures. According to the pressure-temperature table for this gas that can be consulted on the Internet, at a temperature of 80°F, the pressure at which this gas is located is 173 psig.

gregori [183]3 years ago
4 0

Answer:

288 psig

Explanation:

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Which property of a substance can be determined using a pH indicator?
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Recall that you're hypothesis is that these values are the fractions of atoms that are still radioactive after n half life cycle
Sergeu [11.5K]

A= 0.5, B = 0.25 , C = 0.125, D = 0.015625 and E = 0.00390625

<h3>What is a hypothesis?</h3>

A research hypothesis is a statement of expectation or prediction that will be tested by research. Before formulating your research hypothesis, read about the topic of interest to you.

Half life of a substance is defined as the amount of time taken by the substance to reduce to half of its original amount.

Here n represents the number of half lives.

The amount of substance that remains after n half lives can be calculated using the given formula,

So when we have n =1,

Fraction of substance that remains = 0.5¹ = 0.5.

That means after first half life over, the amount of substance that remains is 0.5 times that of original.

Therefore we have A = 0.5

When n = 2, we have 0.5² = 0.25

So when 2 half lives are over, the amount of substance that remains is 0.25 times that of original

Therefore B = 0.25

When n = 3, we have 0.5³ = 0.125

So when 3 half lives are over, the amount of substance that remains is 0.125 times that of original.

Therefore we have C = 0.125

When n = 6 , we have 0.5⁶ = 0.015625

So D = 0.015625

When n = 8, we have 0.5⁸ = 0.00390625

Therefore E = 0.00390625

The values for A, B, C, D and E are 0.5, 0.25, 0.125, 0.015625 and 0.00390625 respectively.

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4 0
2 years ago
3 Mg + 1 Fe2O3 --&gt; 2 Fe + 3 MgO
Gala2k [10]

Answer:

Fe₂O₃ is the limiting reactant.

7.57 g of MgO are formed.

Explanation:

  • 3 Mg + 1 Fe₂O₃ → 2 Fe + 3 MgO

First we <u>convert the given masses of both reactants into moles</u>, using their <em>respective molar masses</em>:

  • 15.6 g Mg ÷ 24.305 g/mol = 0.642 mol Mg
  • 10.0 g Fe₂O₃ ÷ 159.69 g/mol = 0.0626 mol Fe₂O₃

0.0626 moles of Fe₂O₃ would react completely with (3 * 0.0626 ) 0.188 moles of Mg. As there are more Mg moles than required, Mg is the reactant in excess; thus, <em>Fe₂O₃ is the limiting reactant</em>.

We now <u>calculate how many MgO moles are produced</u>, using the <em>number of moles of the limiting reactant</em>:

  • 0.0626 mol Fe₂O₃ * \frac{3molMgO}{1molFe_2O_3} = 0.188 mol MgO

Finally we <u>convert moles of MgO into grams</u>:

  • 0.188 mol MgO * 40.3 g/mol = 7.57 g
8 0
3 years ago
If a sample of oxygen gas originally at from 171.4 K has its temperature increased to 288.4 K and
julia-pushkina [17]

Answer:

V₂ = 0.95 L

Explanation:

Given data:

Initial temperature of gas = 171.4 K

Final temperature of gas = 288.4 K

Final volume = 1.6 L

Initial volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₁ = V₂T₁ /T₂

V₂ = 1.6 L × 171.4 K / 288.4 k

V₂ = 274.24 L.K / 288.4 K

V₂ = 0.95 L

6 0
3 years ago
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