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2 years ago

Why krypton and noble gas are not used in airships using the periodic table?

Chemistry

1 answer:

WITCHER [35]2 years ago

3 0

Answer:

Since the noble gases are unreactive or inert, they are safe to use. Helium is used to fill balloons and airships, because it is much lighter that air and it will not catch fire. Neon is used in advertising signs. It will give red glow, but the color can be changes by mixing it with other gases.

Explanation:

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vodomira [7]

Answer:

Final volume 30.513 L.

Explanation:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Given data:

Initial volume = 17 L

Initial pressure = 2.3 atm

Initial temperature = 299 K

Final temperature = 350 K

Final volume = ?

Final pressure = 1.5 atm

Solution:

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁ T₂/ T₁ P₂

V₂ = 2.3 atm × 17 L × 350 K / 299 K × 1.5 atm

V₂ = 13685 atm .L. K / 448.5 K . atm

V₂ = 30.513 L

6 0

2 years ago

Number 6 Yususueudidididifididi

Furkat [3]

The answer is predict, because it fits, and because someone can predict the outcome of something.

6 0

3 years ago

Which of these is a chemical property?

ankoles [38]

Answer:

The second choice, or flammability.

Explanation:

The flammability of something is how easy it is for it to burn or ignite.

7 0

3 years ago

Read 2 more answers

g Ethanol boils at 78.4 °C with \DeltaΔHvap = 38.6 kJ/mol. A 0.200-mol sample of ethanol is heated from some colder temperature

andrezito [222]

Answer:

8.77 kilo Joules will be the total amount of heat required for both the heating and the vaporizing.

Explanation:

Moles of ethanol of ethanol = 0.200 mol

Heat required to heat 0.200 moles of ethanol = Q = 1.05 kJ

Enthalpy of vaporization of ethanol = $\Delta H_{vap}=38.6 kJ/mol$

Heat required to vaporize 0.200 moles of ethanol = Q'

$Q'=\Delta H_{vap}\times 0.200 mol=38.6 kJ/mol\times 0.200 mol=7.72 kJ$

Total heat required to fore heating and the vaporizing :

= Q + Q' = 1.05 kJ + 7.72 kJ = 8.77 kJ

8.77 kilo Joules will be the total amount of heat required for both the heating and the vaporizing.

7 0

3 years ago

2. Answer the following questions about a sample of calcium phosphate:

weqwewe [10]

Answer:

a) 310.18 g/mol

b) 4.352 moles Ca3(PO4)2

c) 2.6 * 10^24 molecules

d) 5.24 * 10^24 P atoms

e)13.056 moles Ca

f)10825.3 grams

Explanation:

Step 1: Data given

Atomic mass of Ca = 40.08 g/mol

Atomic mass of P = 30.97 g/mol

Atomic mass of O = 16.0 g/mol

Step 2: Calculate molecular weight of Ca3(PO4)2

Molecular weight of Ca3(PO4)2 = 3*atomic mass of Ca + 2* atomic mass of P and 8* atomic mass of O

Molecular weight of Ca3(PO4)2 = 3*40.08 + 2*30.97 + 8*16.0 = 310.18 g/mol

Step 3: Calculate moles of Ca3(PO4)2 in 1350 grams

Moles Ca3(PO4)2 = mass Ca3(PO4)2 /molar mass

Moles Ca3(PO4)2 = 1350 grams / 310.18 g/mol

Moles Ca3(PO4)2 = 4.352 moles

Step 4: Calculate molecules in 1350 grams

Molecules = moles * number of Avogadro

Molecules = 4.352 moles * 6.02 * 10^23

Molecules = 2.6 *10^24 molecules

Step 5: Calculate moles Phosphorus

For 1 mol Ca3(PO4)2 we need 2 moles P

For 4.352 moles Ca3(PO4)2 we have 2*4.352 = 8.704 moles

Step 6: Calculate P atoms

Atoms P = 8.704 moles * 6.02*10^23

Atoms P = 5.24 * 10^24 P atoms

Step 7: Calculate moles Calcium in 1350 grams

For 1 mol Ca3(PO4)2 we have 3 moles Ca

For 4.352 moles we have 3*4.352 = 13.056 moles Ca

Step 8: Calculate mass of 2.1 * 10^25 molecules of Ca3(PO4)2

Moles Ca3(PO4)2 = 2.1 * 10^25 / 6.02 * 10^23

Moles Ca3(PO4)2 = 34.9 moles

Mass Ca3(PO4)2 = 34.9 moles * 310.18 g/mol

Mass Ca3(PO4)2 = 10825.3 grams

4 0

2 years ago