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Firdavs [7]
3 years ago
10

Explain why gases such as the oxygen found in tanks used as hospitals are compressed. why must care be taken to prevent compress

ed gases
Chemistry
2 answers:
allochka39001 [22]3 years ago
7 0

They are compressed so that a larger amount of gas can be stored in a smaller container. A greater mass confined to a smaller volume makes transporting and storing of gases easier. Increasing temperature increases pressure, and the cylinders might explode. Before compressed oxygen can be breathed, it must be decompressed.

diamong [38]3 years ago
7 0

Answer:  Large Amount of gas can be stored in the small container

Explanation:  Gases such as oxygen found in tanks used as hospitals are compressed because on compressing the gas, its volume decreases which will help to store large amount of gas.

This large amount can be stored in the small container.

Care must be taken as temperature should be not increased because as temperature will increase, pressure will also increase and this could lead to the explosion of the cylinder . Thus preventive measures must be taken.

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How many atoms are centered on the (100) plane for the fcc crystal structure?
Setler [38]

Here, we have to get the number of atoms present in the 100 plane of the FCC crystal lattice.

There will be 2 atoms in 100 plane of FCC crystal lattice.

In the face centered crystal (FCC) lattice there are atoms at each corner of the cube and each are shared by 4 another atoms. And an atom is present at the face of the crystal.

For the 100 plane of the Miller indices the intercepts are a, ∞, ∞ or 2a, ∞, ∞.

Thus, for the 4 atoms of the corner at the cube shared by 4 other atoms will contribute, 4 × \frac{1}{4} = 1 and the un-shared atoms at the face will contribute another 1, which make the total atom 1 + 1 = 2.

5 0
2 years ago
Which best describes an element? A pure substance. A type of a mixture. A pure compound. An impure substance.
mr Goodwill [35]

a pure compound because an element is untouched and is just itself

7 0
3 years ago
Read 2 more answers
What are the products of the neutralization reaction between HCl and
liraira [26]

Answer:

opinion b

Explanation:

the product of neutralization reaction between hcl and CA(oh)2 is option b.

4 0
2 years ago
How are distillation and reverse osmosis alike?
Andrei [34K]
I'm having trouble with this one. At the very least, they both purify water. If I'm not mistaken, distillation is part of the process of purifying water.
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5 0
3 years ago
Find percent yield:
saveliy_v [14]

<u>Answer:</u> The percent yield of the reaction is 91.8 %

<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)

  • <u>For B_5H_9 :</u>

Given mass of B_5H_9 = 4.0 g

Molar mass of B_5H_9 = 63.12 g/mol

Putting values in equation 1, we get:

\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol

The chemical equation for the reaction of B_5H_9 and oxygen gas follows:

2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of B_2H_5

So, 0.3125 moles of oxygen gas will react with = \frac{2}{12}\times 0.3125=0.052mol of B_2H_5

As, given amount of B_2H_5 is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of B_2O_3

So, 0.3125 moles of oxygen gas will produce = \frac{5}{12}\times 0.3125=0.130moles of water

Now, calculating the mass of B_2O_3 from equation 1, we get:

Molar mass of B_2O_3 = 69.93 g/mol

Moles of B_2O_3 = 0.130 moles

Putting values in equation 1, we get:

0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g

To calculate the percentage yield of B_2O_3, we use the equation:

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

Experimental yield of B_2O_3 = 8.32 g

Theoretical yield of B_2O_3 = 9.052 g

Putting values in above equation, we get:

\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%

Hence, the percent yield of the reaction is 91.8 %

6 0
3 years ago
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