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juin [17]
3 years ago
15

Trimix is a general name for a type of gas blend used by technical divers and contains nitrogen, oxygen and helium. In one Trimi

x blend, the partial pressures of each gas are 55 atm oxygen, 90 atm nitrogen, and 50 atm helium. What is the percent oxygen (by volume) in this trimex blend?
Chemistry
2 answers:
Keith_Richards [23]3 years ago
6 0

Answer:

28.2

Explanation:

Add all of the pressures, 55, 90, and 50, and divide 100 by the answer you get (195). You'll get 0.512820513 and multiply it by .55 (atm of Oxygen) and you'll get 28.2

Korolek [52]3 years ago
3 0

Answer:

The percentage by volume of Oxygen gas in the Trimex blend is 28.2%.

Explanation:

We shall use the formula that connects the mole ratio of a particular gas in a mixture with the partial pressure of that gas and the total pressure of the mixture: which is:

Mole ratio of Oxygen in Trimex = Partial pressure of Oxygen in Trimex divided by the Total pressure of all the gases that made up of Trimex.

Here we go, mole ratio of Oxygen= 55/195. The total pressure is the addition of the partial pressures of all the gases in the Trimex mixture, which is 55+90+50=195.

So the mole ratio of our Oxygen gas will be: 0.282, no unit.

This means we can assume the total moles of all the gas is 1, and oxygen is 0.282.

Since we know that 1 mole of any gas will have 22.4dm cube as volume, then the total volume of the Trimex will be 22.4dm cube, while that of the oxygen gas will be 0.282 X 22.4 dm cube= 6.3179dm cube. Therefore, the percentage by volume of Oxygen gas in the Trimex will be: 6.3179/22.4 X 100 = 28.2%.

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If a compound has two atoms of aluminum (AI) and three atoms of oxygen (O) what would its chemical formula look like?
liberstina [14]

Answer:

Al₂O₃

Explanation:

that's the molecular formula for aluminum oxide/alumina.

3 0
3 years ago
Chromium(III) oxide can be prepared by heating chromium(IV) oxide in vacuo at high temperature: 4Cr02 —2Cr2O3 +02 The reaction o
kkurt [141]

<u>Answer:</u> The theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of CrO_2 = 480.1 g

Molar mass of CrO_2 = 84 g/mol

Putting values in equation 1, we get:

\text{Moles of }CrO_2=\frac{480.1g}{84g/mol}=5.72mol

For the given chemical equation:

4CrO_2\rightarrow 2Cr_2O_3+O_2

By Stoichiometry of the reaction:

4 moles of CrO_2 produces 2 moles of chromium (III) oxide

So, 5.72 moles of CrO_2 will produce = \frac{2}{4}\times 5.72=2.86mol of chromium (III) oxide

Now, calculating the mass of chromium (III) oxide from equation 1, we get:

Molar mass of chromium (III) oxide = 152 g/mol

Moles of chromium (III) oxide = 2.86 moles

Putting values in equation 1, we get:

2.86mol=\frac{\text{Mass of chromium (III) oxide}}{152g/mol}\\\\\text{Mass of chromium (III) oxide}=(2.86mol\times 152g/mol)=434.72g

To calculate the percentage yield of chromium (III) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of chromium (III) oxide = 402.4 g

Theoretical yield of chromium (III) oxide = 434.72 g

Putting values in above equation, we get:

\%\text{ yield of chromium (III) oxide}=\frac{402.4g}{434.72g}\times 100\\\\\% \text{yield of chromium (III) oxide}=\%

Hence, the theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

7 0
3 years ago
Indicate which of the following criteria are important for the selection of a buffer to use in an in vitro biochemical reaction.
JulsSmile [24]

Answer:

The correct answer is: d. The pKa of the chosen buffer should be close to the optimal pH for the biochemical reaction.

Explanation:

The buffer resist or maintain the change in pH in case of Acid or basic addition to the solution. The buffer capacity should be within one or two pH units when compared to the optimal pH.

Thus it is important to select a buffer with pKa close to the optimum pH of the reaction because the ability for the buffer to maintain the pH is is great at the pH close to pKa.

7 0
3 years ago
As part of a science experiment, Jose did a test for starch on a slice of apple and a slice of potato. The yellow-orange iodine
juin [17]

Answer:

See explanation

Explanation:

Iodine solution is used to test for starch.  A positive test for starch gives a blue-black color.

The fact that the color of the apple remained the same is indicative of the fact that starch was not contained in the apple.

A change in the color of potato indicates the presence of starch in the potato.

The fact that iodine did not react with apple should not be taken to mean that apples contain no starch at all. Starch changes gradually to sugar as fruits ripen. This is why the apple gave a negative test for starch.

5 0
3 years ago
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Andrews [41]

<u>Answer:</u> The experimental van't Hoff factor is 1.21

<u>Explanation:</u>

The expression for the depression in freezing point is given as:

\Delta T_f=iK_f\times m

where,

i = van't Hoff factor = ?

\Delta T_f = depression in freezing point  = 0.225°C

K_f = Cryoscopic constant  = 1.86°C/m

m = molality of the solution = 0.100 m

Putting values in above equation, we get:

0.225^oC=i\times 1.86^oC/m\times 0.100m\\\\i=\frac{0.225}{1.86\times 0.100}=1.21

Hence, the experimental van't Hoff factor is 1.21

7 0
3 years ago
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