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horrorfan [7]
3 years ago
6

1. When the temperature of a rigid hollow sphere containing 685 L of helium gas is

Chemistry
1 answer:
mote1985 [20]3 years ago
6 0
<h3>Answer:</h3>

250.756 moles He

<h3>Explanation:</h3>

From the question we are given;

Volume, L = 685 L

Temperature, T = 621 K

Pressure, P = 189 × 10 kPa

We are required to calculate the number of moles of the gas,

Using the Ideal gas equation,

PV = nRT, where P is the pressure, V is the volume, T is the temperature, n is the number of moles, and R is the ideal gas constant.

We can replace the known variables and constant in the equation to get the unknown variable, n.

Using ideal gas constant as 8.3145 L.kPa/K/mol

n=\frac{PV}{RT}

n=\frac{(1890)(685)}{(8.3145)(621)}

n = 250.7557 moles

   n = 250.756 moles

The moles of helium contained in the sphere is 250.756 moles

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b-1 CO2+ 2H2O

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4 years ago
In the lab you measure a clean dry crucible and cover to be 24.36 grams. You obtain a 2cm piece of pure magnesium metal. After s
mart [117]

Answer:

1) 0.3g Mg

2)0.5g MgO

3)0.2g O

4)0.01mol Mg & 0.01mol O

5)0.01mol MgO

6) Empirical formula MgO

Explanation:

The mass og Mg is obtained by substracting 24.36g from 24.66g:

24.66 - 24.36 = 0.3g Mg

The ignition of Mg means that it's reacting with oxygen to form an oxide. The increase in the crucible mass after the Mg ignition is due to the addition of oxygen. However, the addition of few drops of water produces a new compound: a hydroxide. According to the oxidation state og Mg (2+), the only magnesium oxide possible is MgO. It happens because the oxidation state of oxygen in oxides is 2-. Which means that just one oxygen atom is required to electrically neutralize one magnesium atom.

We can use a conversion factor to know how much MgO is made from from 0.3 g of Mg:

0.3g Mg*\frac{16gO}{24.3gMg}= 0.2g O

Thereby the mass of the oxide is 0.2g O + 0.3g Mg = 0.5g MgO

We convert the mass of oxygen and magnesium to the respective amounts in moles by using conversion factors:

0.2g O*\frac{1 mol O}{16g O}= 0.01mol O

0.3g Mg*\frac{1mol Mg}{24.3g Mg}= 0.01mol Mg

The moles of MgO can be obtained from:

0.5g MgO*\frac{1mol MgO}{40.3g MgO}= 0.01mol MgO

To obtain the empirical formula, the amount fo moles of each elements must be divided by the smallest one, in this case, 0.01.

The result for both number of  Mg atoms and O atoms is 1. This can be interpreted to mean that there is a Mg atom for each O atom forming the  formula unit of the compound.

The step when water is added to the compound resulting after heating does not affect the calculations necessary for the magnesium oxide.

4 0
3 years ago
The activation energy of an uncatalyzed reaction is 95kJ/mol. The addition of a catalyst lowers the activation energy to 55kJ/mo
notka56 [123]

Answer:

a) at 25°C the rate of reaction increases by a factor of 1,027*10^7

b) at 25°C the rate of reaction increases by a factor of 1,777*10^5

Explanation:

using the Arrhenius equation

k= ko*e^(-Ea/RT)

where

k= reaction rate

ko= collision factor

Ea= activation energy

R= ideal gas constant= 8.314 J/mol*K

T= absolute temperature

for the uncatalysed reaction

k1= ko*e^(-Ea1/RT)

for the catalysed reaction

k2= ko*e^(-Ea2/RT)

dividing both equations

k2/k1= e^(-(Ea2-Ea1)/RT)

a) at 25°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,027*10^7

therefore at 25°C , k2/k1 = 1,027*10^6

b) at 125°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,777*10^5

therefore at 125°C , k2/k1 = 1,777*10^5

Note:

when the catalysts is incorporated, the catalysed reaction and the uncatalysed one run in parallel and therefore the real reaction rate is

k real = k1 + k2 = k2 (1+k1/k2)

since k2>>k1 → 1+k1/k2 ≈ 1 and thus k real ≈ k2

6 0
3 years ago
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