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tresset_1 [31]
2 years ago
12

How many moles of N2 are need to fill a 35 L tank at standard temperature and pressure?

Chemistry
1 answer:
Tems11 [23]2 years ago
4 0

1.56 moles of N2 are needed to fill a 35 L tank at standard temperature and pressure. Details about moles can be found below.

<h3>How to calculate number of moles?</h3>

The number of moles of a substance can be calculated using the following formula:

PV = nRT

Where;

  • P = pressure
  • V = volume
  • n = number of moles
  • R = gas law constant
  • T = temperature

At STP;

  • T = 273K
  • P = 1 atm
  • R = 0.0821 Latm/molK

1 × 35 = n × 0.0821 × 273

35 = 22.41n

n = 35/22.41

n = 1.56mol

Therefore, 1.56 moles of N2 are needed to fill a 35 L tank at standard temperature and pressure.

Learn more about number of moles at: brainly.com/question/14919968

#SPJ1

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A 225 g sample of an unknown solid is heated 67C and placed into a calorimeter containing 25.6 g of water at 15.6°c. If the fina
Stella [2.4K]

Answer:

= 1.271 J/g°C

Explanation:

Heat released by the metal sample will be equivalent to the heat absorbed by  water.

But heat = mass × specific heat capacity × temperature change

Thus;

Heat released by the solid;

= 225 g × c ×(67 -53) , where c is the specific heat capacity of the metal

= 3150 c joules

Heat absorbed by water;

= 25.6 g × 4.18 J/g°C × (53-15.6)

= 4002.0992  joules

Therefore;

3150 c joules = 4002.0992 joules

c =4002.0992/3150

 <u> = 1.271 J/g°C</u>

6 0
3 years ago
What is the wavelength of the matter wave associated with an electron (me= 9.1 x 10-31 kg) moving with a speed of 2.5 x 107 m/s?
riadik2000 [5.3K]
And h value is constant 6.624 x 10^-34

3 0
3 years ago
2c4H10 +13O2 —&gt;8CO2+10H2O to find how many moles of oxygen would react with 4.3 mol C4H10
Elena-2011 [213]

Answer:

\large\boxed{\text{28 mol of O$_{2}$}}

Explanation:

             2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

n/mol:      4.3  

13 mol of O₂ react with 2 mol of 2C₄H₁₀

\text{Moles of O}_{2} = \text{4.3 mol C$_{4}$H$_{10}$} \times \dfrac{\text{13 mol O}_{2}}{\text{2 mol C$_{4}$H$_{10}$}} = \textbf{28 mol O}_{2}\\\\\text{The reaction requires $\large\boxed{\textbf{28 mol of O$_{2}$}}$}

5 0
3 years ago
Arrange the following H atom electron transitions in order of decreasing wavelength of the photon absorbed or emitted:
Katyanochek1 [597]

The decreasing order of wavelengths of the photons emitted or absorbed by the H atom is : b → c → a → d

Rydberg's formula :

                                   \frac{1}{\lambda} = R_h (\frac{1}{n_1^2}-\frac{1}{n_2^2} ),

where  λ is the wavelength of the photon emitted or absorbed from an H atom electron transition from n_1 to n_2 and R_h = 109677 is the Rydberg Constant. Here n_1 and  n_2 represents the transitions.

(a) n_1 =2 to n_2 = infinity

            \frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{\infty^2}  ) = 109677/4     [since 1/infinity = 0] Therefore, \lambda = 4 / 109677 = 0.00003647 m

(b) n_1=4 to  n_2 = 20

           \frac{1}{\lambda} = 109677\times (\frac{1}{4^2}-\frac{1}{20^2}  ) = 6580.62

Therefore,  \lambda = 1 / 6580.62 = 0.000152 m

(c) n_1=3 to  n_2 = 10

          \frac{1}{\lambda} = 109677\times (\frac{1}{3^2}-\frac{1}{10^2}  ) = 11089.56

Therefore,  \lambda = 1 / 11089.56 = 0.00009 m

(d)  n_1=2 to  n_2 = 1

          \frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{1^2}  ) = - 82257.75

Therefore,  \lambda = 1 /82257.75  = - 0.0000121 m  

[Even though there is a negative sign, the magnitude is only considered because the sign denotes that energy is emitted.]

So the decreasing order of wavelength of the photon absorbed or emitted is b → c → a → d.

Learn more about the Rydberg's formula athttps://brainly.com/question/14649374

#SPJ4

8 0
1 year ago
1/4 checkpoint
DochEvi [55]

Answer:

here's the answers

Explanation:

7 0
2 years ago
Read 2 more answers
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