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KiRa [710]
3 years ago
8

A pressure cooker contains 5.68 L of air at a temperature of 390 4K if the absolute pressure of the air in the pressure cooker i

s 205 Pa how many moles of air are in the cooker
Chemistry
1 answer:
Cloud [144]3 years ago
6 0

Answer:

3.59x10⁻⁴ mol

Explanation:

Assuming ideal behaviour we can solve this problem by using the<em> PV=nRT formula</em>, where:

  • P = 205 Pa
  • V = 5.68 L
  • n = ?
  • R = 8314.46 Pa·L·mol⁻¹·K⁻¹
  • T = 390.4 K

We<u> input the data given by the problem</u>:

  • 205  Pa * 5.68 L = n * 8314.46 Pa·L·mol⁻¹·K⁻¹ * 390.4 K

And <u>solve for n</u>:

  • n = 3.59x10⁻⁴ mol
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NikAS [45]

Answer:

Solar energy is a renewable free source of energy

Explanation:

It is sustainable and totally inexhaustible, unlike fossil fuels which are finite. It is also a non-polluting source of energy and it does not emit any greenhouse gases when producing electricity.

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2 years ago
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An aqueous solution in a 55 gallon (208 l drum), characterized by minimal buffering capacity, received 4kg of phenol and 1.5 kg
lianna [129]

An aqueous solution in a 55 gallon (208 l drum), characterized by minimal buffering capacity, received 4kg of phenol and 1.5 kg of sodium phenate. What is the ph of the solution. The pka of phenol = 9.98. Mw of phenol and sodium phenate are 94 g/mol and 116 g/mol, respectively.


Volume of solution = 55 gallons = 208.2 L [ 1 gallon = 3.78 L]

moles of phenol = mass / molar mass = 4000 g / 94 = 42.55 moles

moles of sodium phenate = mass / molar mass = 1500 / 116 = 12.93 moles

pKa of phenol = 9.98

We know that the pH of buffer is calculated using Hendersen Hassalbalch's equation

pH = pKa + log [salt] / [acid]

volume is same for both the sodium phenate and phenol has we can directly take the moles of each in the formula

pH = 9.98 + log [12.93 / 42.55] = 9.46



4 0
3 years ago
Polyatomic ion formula for aluminum hydroxide
KonstantinChe [14]

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3 years ago
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You need to prepare 150 mL of 0.1 M solution of silver chloride. How much silver chloride is required?
Rudiy27

Answer:

amount of silver chloride required is 0.015 moles or 2.1504 g

Explanation:

0.1M AgCL means 0.1mol/dm³ or 0.1mol/L

1L = 1000mL

if 0.1mol of AgCl is contained in 1000mL of solution

then x will be contained in 150mL of solution

cross multiply to find x

x = (0.1*150)/1000

x= 0.015 moles

moles of silver chloride present in 150 mL of solution is 0.15 moles

To convert this to grams, simply multiply this value by the molar mass of silver chloride

molar mass of silver chloride AgCl =107.86 + 35.5

                                                     =143.36 g/mol

mass of AgCl = moles *molar mass

                       =0.015*143.36

                        =2.1504g

                        =

4 0
3 years ago
Hydrogen is manufactured on an industrial scale by this sequence of reactions: Write an equation that gives the overall equilibr
RideAnS [48]

The question is incomplete. The complete question is :

Hydrogen is manufactured on an industrial scale by this sequence of reactions:

$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g    ) \ \ \ \ \ \ \ \ \ \ K_1$

$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \ \ \ \ \ \ \ \ \ \ \ \  K_2$

The net reaction is  :

$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g) \ \ \ \ \ \ \ \ \ K$

Write an equation that gives the overall equilibrium constant K in terms of the equilibrium constants K_1 and K_2. If you need to include any physical constants, be sure you use their standard symbols, which you'll find in the ALEKS Calculator.

Solution :

$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g    ) \ \ \ \ \ \ \ \ \ \ K_1$

$K_1 = \frac{[CO][H_2]^3}{[CH_4][H_2O]}$     ...............(1)

$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \ \ \ \ \ \ \ \ \ \ \ \  K_2$

$K_2 = \frac{[CO_2][H_2]}{[CO][H_2O]}$  ...................(2)

$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g) \ \ \ \ \ \ \ \ \ K$

$K=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$

On multiplication of equation (1) and (2), we get

$K_1 \times K_2=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \times \frac{[CO_2][H_2]}{[CO][H_2O]}$

$K_1K_2=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$  .................(4)

Comparing equation (3) and equation (4), we get

$K=K_1K_2$

4 0
2 years ago
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