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Vlad1618 [11]
2 years ago
6

A balloon is filled to a volume of 18.9L at a temperature of 27.1ºC. If the pressure in the balloon is measured to be 2.20 atm,

how many moles of gas are contained inside the balloon?
Chemistry
1 answer:
ryzh [129]2 years ago
4 0

<u>Answer:</u> The number of moles of gas in balloon is 1.69 moles

<u>Explanation:</u>

To calculate the number of moles of gas, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 2.20 atm

V = Volume of the gas = 18.9 L

T = Temperature of the gas = 27.1^oC=[27.1+273]=300.1K

R = Gas constant = 0.0821\text{ L. atm }mol^{-1}K^{-1}

n = number of moles of gas = ?

Putting values in above equation, we get:

2.20atm\times 18.9L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 300.1K\\\\n=\frac{2.20\times 18.9}{0.0821\times 300.1}=1.69mol

Hence, the number of moles of gas in balloon is 1.69 moles

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A 1300 mL sample of gas with a molar mass of 71.0 g/mol at STP has what density?
Blizzard [7]

Answer:

0.055g/mL

Explanation:

Data obtained from the question include:

Molar Mass of the gass sample = 71g/mol

Volume of the gas sample = 1300 mL

Density =?

The density of a substance is simply mass per unit volume. It is represented mathematically as:

Density = Mass /volume.

With the above equation, we can easily obtain the density of sample of gas as illustrated below:

Density = 71g / 1300 mL

Density = 0.055g/mL

Therefore, the density of the gas sample is 0.055g/mL

6 0
3 years ago
If 175mL of oxygen is produced at STP, how many grams of hydrogen peroxide, H2O2
Vlad [161]

Answer:

0.53g

Explanation:

We'll begin by converting 175mL to L. This is illustrated below:

1000mL = 1L

Therefore 175mL = 175/1000 = 0.175L

Next, we shall calculate the number of mole of O2 that occupy 0.175L. This is illustrated below:

1 mole of O2 occupy 22.4L at stp.

Therefore, Xmol of O2 will occupy 0.175L i.e

Xmol of O2 = 0.175/22.4

Xmol of O2 = 7.81×10¯³ mole

Therefore, 7.81×10¯³ mole of O2 occupy 175mL.

Next, we shall determine the number of mole of H2O2 that decomposed to produce 7.81×10¯³ mole of O2. This is illustrated below:

2H2O2 —> 2H2O + O2

From the balanced equation above,

2 moles of H2O2 decomposed to produce 1 mole of O2.

Therefore, Xmol of H2O2 will decompose to produce 7.81×10¯³ mole of O2 i.e

Xmol of H2O2 = 2 x 7.81×10¯³

Xmol of H2O2 = 1.562×10¯² mole

Therefore, 1.562×10¯² mole of H2O2 decomposed in the reaction.

Finally, we shall convert 1.562×10¯² mole of H2O2 to grams. This is illustrated below:

Molar mass of H2O2 = (2x1) + (16x2) = 34g/mol

Mole of H2O2 = 1.562×10¯² mole

Mass of H2O2 =..?

Mole = mass /Molar mass

1.562×10¯² = mass /34

Cross multiply

Mass of H2O2 = 1.562×10¯² x 34

Mass of H2O2 = 0.53g

Therefore, 0.53g of Hydrogen peroxide, H2O2 were decomposition in the reaction.

3 0
3 years ago
If you combine 230.0 mL 230.0 mL of water at 25.00 ∘ C 25.00 ∘C and 120.0 mL 120.0 mL of water at 95.00 ∘ C, 95.00 ∘C, what is t
Thepotemich [5.8K]

<u>Answer:</u> The final temperature of the mixture is  49°C

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

  • <u>For cold water:</u>

Density of cold water = 1 g/mL

Volume of cold water = 230.0 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{230.0mL}\\\\\text{Mass of water}=(1g/mL\times 230.0mL)=230g

  • <u>For hot water:</u>

Density of hot water = 1 g/mL

Volume of hot water = 120.0 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{120.0mL}\\\\\text{Mass of water}=(1g/mL\times 120.0mL)=120g

When hot water is mixed with cold water, the amount of heat released by hot water will be equal to the amount of heat absorbed by cold water.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]      ......(1)

where,

q = heat absorbed or released

m_1 = mass of hot water = 120 g

m_2 = mass of cold water = 230 g

T_{final} = final temperature = ?°C

T_1 = initial temperature of hot water = 95°C

T_2 = initial temperature of cold water = 25°C

c = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

120\times 4.186\times (T_{final}-95)=-[230\times 4.186\times (T_{final}-25)]

T_{final}=49^oC

Hence, the final temperature of the mixture is  49°C

4 0
3 years ago
Atmospheric air pressure is measured with a _____.
Svetradugi [14.3K]
It is measured with a barometer
3 0
3 years ago
Read 2 more answers
4. For water, H20 represents
timama [110]

Answer:

The empirical and molecular formula because the ratio is 2:1 <em>and </em>there is 2 hydrogen atoms and 1 oxygen atom.

Hope this helps!

3 0
2 years ago
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