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Anna11 [10]
3 years ago
10

Part 1: A cylinder containing 20.0 L of compressed nitrogen is connected to an empty (evacuated) vessel with an unknown volume.

The gas pressure in the cylinder starts at 25 atm and drops to 2 atm without a change in temperature. Determine the volume of the vessel.Part 2: A mixture of H2 and He are in a 10.0 L vessel at 273 K. The total pressure is 756 torr. What is the partial pressure of H2 in the vessel if XHe = 0.75?
Chemistry
1 answer:
baherus [9]3 years ago
7 0

Answer:

The volume of the vessel is 250 L

Partial pressure of hydrogen = 189 torr

Explanation:

Using Boyle's law  

{P_1}\times {V_1}={P_2}\times {V_2}

Given ,  

V₁ = 20.0 L

V₂ = ?

P₁ = 25 atm

P₂ = 2 atm

Using above equation as:

{P_1}\times {V_1}={P_2}\times {V_2}

{25}\times {20.0}={2}\times {V_2}

{V_2}=\frac {{25}\times {20.0}}{2}\ L

{V_2}=250\ L

<u>The volume of the vessel is 250 L.</u>

According to Dalton's law of partial pressure:-

P_{H_2}=Mole\ fraction\times Total\ Pressure

So, according to definition of mole fraction:

Mole\ fraction\ of\ H_2=\frac {n_{H_2}}{n_{H_2}+n_{He}}

Also,

Mole fraction of H₂ = 1 - Mole fraction of He = 1 - 0.75 = 0.25

So,

Total pressure = 756 torr

Thus,

P_{H_2}=0.25\times 756\ torr

<u>Partial pressure of hydrogen = 189 torr.</u>

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nevsk [136]

Answer:

The volume of carbon dioxide gas generated 468 mL.

Explanation:

The percent by mass of bicarbonate in a certain Alka-Seltzer = 32.5%

Mass of tablet = 3.45 g

Mass of bicarbonate =3.45 g\times \frac{32.5}{100}=1.121 mol

Moles of bicarbonate ion = \frac{1.121 g/mol}{61 g/mol}=0.01840 mol

HCO_3^{-}(aq)+HCl(aq)\rightarrow H_2O(l)+CO_2(g)+Cl^-(aq)

According to reaction, 1 mole of bicarbonate ion gives with 1 mole of carbon dioxide gas , then 0.01840 mole of bicarbonate ion will give:

\frac{1}{1}\times 0.01840 mol=0.01840 mol of carbon dioxide gas

Moles of carbon dioxide gas  n = 0.01840 mol

Pressure of the carbon dioxide gas = P = 1.00 atm

Temperature of the carbon dioxide gas = T = 37°C = 37+273 K=310 K

Volume of the carbon dioxide gas = V

PV=nRT (ideal gas equation)

V=\frac{nRT}{P}=\frac{0.01840 mol\times 0.0821 atm L/mol K\times 310 K}{1.00 atm}=0.468 L

1 L = 1000 mL

0.468 L =0.468 × 1000 mL = 468 mL

The volume of carbon dioxide gas generated 468 mL.

5 0
3 years ago
What are some examples of fusion
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Answer:

Example of a fusion dish: combination of smoked salmon wrapped in rice paper, with avocado, cucumber and crab sticks. Fusion cuisine is cuisine that combines elements of different culinary traditions that originate from different countries, regions, or cultures.

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3 years ago
Krispy Kreme Doughnuts decide to donate $10,000 to a classroom for a party how many donuts did they donate in total
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Find out how much the donuts cost then divide that by $10,000 and you will get your answer

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In the spring of 1984, concern arose over the presence of ethylene dibromide, or EDB, in grains and cereals. EDB has the molecul
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869.6 × 10¹⁴ molecules of EDB

Explanation:

We have 1.9 lb of flour with a EDB concentration of 31.5 ppb.

We need to transform lb in grams.

1 lb = 453.6 grams

1.9 lb = (1.9 × 453.6) / 1 = 861.8 grams

Now we determine the number of molecules of EDB in the sample by devise the following reasoning:

if we have        31.5 × 10⁻⁹ g of EDB in 1 g of sample

then we have   X  g of EDB in 861.8 g of sample

X = (31.5 × 10⁻⁹ × 861.8) / 1 = 27146.7 × 10⁻⁹ g of EDB

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Taking in account that 1 mole of any substance contains 6.022 × 10²³ (Avogadro’s number) molecules we devise the following reasoning:

if       188 g of EDB contains 6.022 × 10²³ molecules

then 27146.7 × 10⁻⁹ g of EDB contains Y molecules

Y = (27146.7 × 10⁻⁹ × 6.022 × 10²³) / 188 = 869.6 × 10¹⁴ molecules of EDB

Learn more:

about Avogadro’s number

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#learnwithBrainly

8 0
3 years ago
Read 2 more answers
Can anybody answer this question of chemistry?
maw [93]

Answer:

Answer:A

Answer:AExplanation:

Answer:AExplanation:Molar Mass of glucose = (6×12)+(1×12)+(16×6)= 180g/mol

= 180g/molNumber of moles of Glucose = Mass/Molar Mass

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778moles

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556moles

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556molesMass of ethanol= Molar Mass of ethanol × Number of moles

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556molesMass of ethanol= Molar Mass of ethanol × Number of moles={(12×2)+(1×6)+16} × 55.5556

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556molesMass of ethanol= Molar Mass of ethanol × Number of moles={(12×2)+(1×6)+16} × 55.5556= 46.5×55.5556

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556molesMass of ethanol= Molar Mass of ethanol × Number of moles={(12×2)+(1×6)+16} × 55.5556= 46.5×55.5556= 2555.55

4 0
2 years ago
Read 2 more answers
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