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Masja [62]
2 years ago
13

Assuming ideal behavior, how many moles of argon would you need to fill a 14.0×12.0×10.0 ft room? assume atmospheric pressure of

1.00 atm, a room temperature of 20.0 ∘c, and 28.2 l/ft3. express your answer with the appropriate units.
Chemistry
1 answer:
Blababa [14]2 years ago
4 0

We use the formula:

PV = nRT

First let us get the volume V:

volume = 14 ft * 12 ft * 10 ft = 1,680 ft^3

Convert this to m^3:

volume = 1680 ft^3 * (1 m / 3.28 ft)^3 = 47.61 m^3

 

n = PV / RT

n = (1 atm) (47.61 m^3) / (293.15 K * 8.21x10^-5 m3 atm / mol K)

<span>n = 1,978.13 mol</span>

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The temperature of 6.24 l of a gas is increased from 25.0°c to 55.0°c at constant pressure. the new volume of the gas is
bixtya [17]
At constant pressure, temperature is directly proportional to the volume and vise versa. The formula will be

V1/T1 = V2/T2

where V1 and T1 are the initial volume and temperature and V2 and T2 are the final volume and temperature. The temperature is in Kelvin and to convert Celsius to Kelvin add 273.

so, 6.24L/298 = V2/328
      =6.87L
6 0
2 years ago
What is the melting point of a solution in which 3.5 grams of sodium chloride is added to 230 mL of water?
kirza4 [7]
We are going to use this equation:

ΔT = - i m Kf

when m is the molality of a solution 

i = 2

and ΔT is the change in melting point = T2- 0 °C

and Kf is cryoscopic constant = 1.86C/m

now we need to calculate the molality so we have to get the moles of NaCl first:

moles of NaCl = mass / molar mass

                         = 3.5 g / 58.44 

                         = 0.0599 moles


when the density of water = 1 g / mL and the volume =230 L

∴ the mass of water = 1 g * 230 mL = 230 g = 0.23Kg 

now we can get the molality = moles NaCl / Kg water

                                                =0.0599moles/0.23Kg

                                                = 0.26 m

∴T2-0 =  - 2 * 0.26 *1.86

∴T2 = -0.967 °C
8 0
2 years ago
Read 2 more answers
A 2.80 g sample of Al reacts with 4.15 g sample of Cl2 according to the equation shown below.
solong [7]

Answer:

Mass = 5.33 g

Explanation:

Given data:

Mass of Al = 2.80 g

Mass of Cl₂ = 4.15 g

Theoretical yield of AlCl₃ = ?

Solution:

Chemical equation:

2Al  +  3Cl₂        →       2AlCl₃

Number of moles of Al:

Number of moles = mass/molar mass

Number of moles = 2.80 g/ 27 g/mol

Number of moles = 0.10 mol

Number of moles of Cl₂:

Number of moles = mass/molar mass

Number of moles = 4.15 g/71 g/mol

Number of moles = 0.06 mol

Now we will compare the moles of AlCl₃ with Al and Cl₂.

                    Cl₂           :        AlCl₃

                    3              :          2

                   0.06         :        2/3×0.06 = 0.04

                   Al             :        AlCl₃

                     2            :          2

                   0.10         :        0.10

Number of moles of AlCl₃ produced by chlorine are less so it will be limiting reactant.

Mass of AlCl₃:Theoretical yield

Mass = number of moles ×molar mass

Mass = 0.04 mol × 133.34 g/mol

Mass = 5.33 g

4 0
1 year ago
Calculate the specific heat capacity for a 22.7-g sample of lead that absorbs 237 J when its temperature increases from 29.8 °C
soldier1979 [14.2K]

Answer:

\boxed {\boxed {\sf c\approx 0.159 \ J/ g \textdegree C}}

Explanation:

We are asked to find the specific heat capacity of a sample of lead. The formula for calculating the specific heat capacity is:

c= \frac{Q}{m \times \Delta T}

The heat absorbed (Q) is 237 Joules. The mass of the lead sample (m) is 22.7 grams. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature. The temperature increases <em>from</em> 29.8 °C <em>to </em>95.6 °C.

  • ΔT = final temperature -inital temperature
  • ΔT= 95.6 °C - 29.8 °C = 65.8 °C

Now we know all three variables and can substitute them into the formula.

  • Q= 237 J
  • m= 22.7 g
  • ΔT = 65.8 °C

c= \frac {237 \ J}{22.7 \ g  \ \times  \ 65.8 \textdegree C}

Solve the denominator.

  • 22.7 g * 65.8 °C = 1493.66 g °C

c= \frac {237 \  J}{1493.66 \ g \textdegree C}

Divide.

c= 0.1586706479 J /g \textdegree C

The original values of heat, temperature, and mass all have 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place tells us to round the 8 up to a 9.

c \approx 0.159 \ J/g \textdegree C

The specific heat capacity of lead is approximately <u>0.159 Joules per gram degree Celsius.</u>

3 0
1 year ago
The energy difference between the 5d and 6s sublevels in gold accounts for its color. Assuming this energy difference is about 2
sergejj [24]

Answer:

\lambda=459.1\times 10^{-7}\ m = 459.1 nm

This wavelength corresponds to yellow color and thus gold has warm yellow color.

Explanation:

Given that:- Energy = 2.7 eV

Energy in eV can be converted to energy in J as:

1 eV = 1.602 × 10⁻¹⁹ J

So, Energy = 2.7\times 1.602\times 10^{-19}\ J=4.33\times 10^{-19}\ J

Considering:-

E=\frac{h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

\lambda is the wavelength of the light

So,  

4.33\times 10^{-19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{\lambda}

4.33\times \:10^{26}\times \lambda=1.99\times 10^{20}

\lambda=459.1\times 10^{-7}\ m = 459.1 nm

This wavelength corresponds to yellow color and thus gold has warm yellow color.

7 0
2 years ago
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