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2 years ago

What is the volume of 0.5 moles of a gas at 2 atmospheres of pressure and 15°c

Chemistry

1 answer:

liq [111]2 years ago

4 0

Answer:

Explanation: convert temperature to kelvin by adding 273 and then input the values into the formula with the given constant

2*v=0.5*0.8206*288 then divide both sides by 2 and get the amount in litres which is 6

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2 years ago

A Downs cell operating at 77.0 A produces 31.0 kg of Na.(a) What volume of Cl₂(g) is produced at 1.0 atm and 540.°C?

nignag [31]

Volume of Cl₂(g) is produced at 1.0 atm and 540.°C=4.5×10^4 L

As per the evenly distributed response

2NaCl (g) ----> 2Na(l)+ Cl2(g)

Calculate the amount of Cl2 that was formed as indicated below:

Moles of Cl2 = 31.0 kg of Na x (1000* 1 * 1 / 1*23* 2)

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P is equal to 1.0 atm, and T is equal to 813.15 K

when converted to Kelvin by multiplying by a factor of 273.15.

Using Cl2 as an ideal gas, determine the in the following volume:

volume = nRT/P

= 673.9 * 0.0821 * 813.15/ 1

=4.5×10^4 L

As a result, the volume of Cl2 under the given circumstances =4.5×10^4 L

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1 year ago

The product side of a chemical reaction is shown.

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The number that represents the coefficient on the product side of the chemical reaction, $-- > 7Ti_2(SO_4)_3$ is 7.

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In equations representing chemical reactions, the coefficient of each reactant or product of a reaction is the number that comes on the left-hand side just before the chemical formula.

The coefficient of each species in a chemical reaction is obtainable when the equation of the reaction is balanced.

For example, in the following equation: 2A + B = 3C + D

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1 year ago

Is potassium a compound?

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Potassium itself is not a compound, it’s an element. Represented by “K” and has an atomic number of 19. However, it can be used to make a compound

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2 years ago

For 2,663 kg of a compound with the formula Al(SO), determine the following quantities (4 pts each); a) The number of moles of t

Nastasia [14]

Answer:

a) 35.485 moles of Al(SO)

b) 35.485 moles of S atoms

c) $2.136197(10^{25})$ Al atoms

d) 567.723 g of O

Explanation:

Let's define the following terms :

1 mol = $6.02.(10^{23})$ elemental units

For example :

1 mol of oxygen atoms = $6.02.(10^{23})$ oxygen atoms

Now, our compound has the following formula

Al(SO)

Where Al is aluminium

S is sulfur

And O is oxygen

All the subscripts are 1 so we can say the following :

1 molecule of Al(SO) has 1 atom of Al , 1 atom of S and 1 atom of O

In terms of moles :

1 mol of Al(SO) has 1 mol of Al , 1 mol of S and 1 mol of O

The molar masses of Al, S and O are

$molarmass_{(Al)}=26.982\frac{g}{mol}$

$molarmass_{(S)}=32.065 \frac{g}{mol}$

$molarmass_{(O)}=15.999\frac{g}{mol}$

If we sum all the molar masses = $(26.982+32.065+15.999)\frac{g}{mol}=75.046\frac{g}{mol}$

Finally, 75.046 g of Al(S0) is 1 mol of Al(SO) which contains 26.982 g of Al, 32.065 g of S and 15.999 g of O.

1 mol of Al(SO) contains 1 mol of Al, 1 mol of S and 1 mol of O.

Now we can calculate a),b),c) and d)

For a)

$2.663 kg=2663g$

75.046 g of Al(SO) = 1 mol of Al(SO)

2663 g of Al(SO) = x

$x=\frac{2663}{75.046}mol=35.485 mol$

2.663 kg of Al(SO) contains 35.485 moles of Al(SO)

b) and c) 1 mol of Al(SO) molecules contains 1 mol of S atoms and 1 mol of Al atoms

We have 35.485 moles of Al(SO) molecules so

We have 35.485 moles of S atoms

And 35.485 moles of Al atoms

If 1 mol = $6.02(10^{23})$

35.485 moles of Al have $(35.485)(6.02)(10^{23})=2.136197(10^{25})$ Al atoms

d) 75.046 g of Al(SO) contains 15.999 g of O

2663 g of Al(SO) contains x g of O

$x=\frac{(2663).(15.999)}{75.046} g$

x = 567.723 g of O

6 0

2 years ago

What is the volume of 0.5 moles of a gas at 2 atmospheres of pressure and 15°c​

What is the volume of 0.5 moles of a gas at 2 atmospheres of pressure and 15°c