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rosijanka [135]
3 years ago
13

If I have 18 liters of gas at a pressure of 88 atm, what will be the final pressure of the gas if the volume of the gas is decre

ased to 12 liters under constant pressure ? Show calculations.
Chemistry
1 answer:
sveticcg [70]3 years ago
6 0

Answer:  You meant in constant temperature?

Because temperature is not mentioned at the beginning.

I suppose temperature remains constant, otherwise

there is no enough data for solving the problem

Explanation: According Boyle´s law in constant temperature

pV = constant.  So p1V1 = p2V2   and pressure p2 = p1V1 /V2

= 88 atm · 18 l / 12 l =  132 atm.

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Which determines the chemical properties of minerals?
azamat
The chemical makeup found on the periodic table.
5 0
3 years ago
At what temperature (in C) will a sample of gas occupy 91.3 L if it occupies 45.0 L at 70.0°C? Assume constant pressure.)
sdas [7]

Solution is here,

for initial case,

temperature(T1)=70°C=70+ 273=343K

vloume( V1) =45 L

for final case,

temperature( T2)=?

volume(V2)= 91.3 L

at constant pressure,

V1/V2 = T1/T2

or, 45/91.3 = 343/ T2

or, T2= (343×91.3)/45

or, T2=695.9 K = (695.9-273)°C=422.9°C

5 0
2 years ago
Identify the group of elements that corresponds to each of the following generalized electron configurations and indicate the nu
Alja [10]

Answer:

(a) [Noble gas] ns² np⁵: Number of unpaired electron is 1 and belongs to group 17 i.e. halogen group of the periodic table.

(b) [noble gas] ns² (n-1)d²: Number of unpaired electron is 2 and belongs to group 4 of the periodic table.

(c) [noble gas] ns² (n-1)d¹⁰ np¹: Number of unpaired electron is 1 and belongs to group 13 of the periodic table.

(d)[noble gas] ns² (n-2)f⁶ : Number of unpaired electron is 6 and belongs to group 8 of the periodic table.

Explanation:

In the Periodic table, the chemical elements are arranged in 7 rows, called periods and 18 columns, called groups. They are organized in increasing order of atomic numbers.

(a) [Noble gas] ns² np⁵

As the total number of electrons in the p-orbital is 5. Therefore, the number of unpaired electron is 1.

This element has 2 electrons in ns orbital and 5 electrons in np orbital. <u>So there are 7 valence electrons.</u>

Therefore, this element belongs to the group 17 i.e. halogen group of the periodic table.

(b) [noble gas] ns² (n-1)d²

As the total number of electrons in the d-orbital is 2. Therefore, the number of unpaired electrons is 2.

This element has 2 electrons in ns orbital and 2 electrons in (n-1)d orbital. So there are <u>4 valence electrons.</u>

Therefore, this element belongs to the group 4 of the periodic table.

(c) [noble gas] ns² (n-1)d¹⁰ np¹

As the total number of electrons in the p-orbital is 1. Therefore, the number of unpaired electron is 1.

This element has 2 electrons in ns orbital and 1 electron in np orbital. So there are <u>3 valence electrons</u>.

Therefore, this element belongs to the group 13 of the periodic table.

(d)[noble gas] ns² (n-2)f⁶

As the total number of electrons in the f-orbital is 6. Therefore, the number of unpaired electron is 6.

This element has 2 electrons in ns orbital and 6 electrons in (n-2)f orbital. So there are <u>8 valence electrons.</u>

Therefore, this element belongs to the group 8 of the periodic table.

3 0
2 years ago
An irregular chunk of iron has a density of 7.87 g/cm3. A 14.5 gram chunk is added to a graduated cylinder. If the final volume
bulgar [2K]

Answer:

37.96 mL

Explanation:

From the question given above, the following data were obtained:

Density (D) of iron = 7.87 g/cm³

Mass (m) of iron = 14.5 g

Volume of Cylinder + Iron = 39.8 mL

Volume of Cylinder =?

Next, we shall determine the volume of the chunk of iron. This can be obtained as follow:

Density (D) of iron = 7.87 g/cm³

Mass (m) of iron = 14.5 g

Volume (V) of iron =?

D = m/V

7.87 = 14.5 /V

Cross multiply

7.87 × V = 14.5

Divide both side by 7.87

V = 14.5 / 7.87

V = 1.84 cm³

Recall:

1 cm³ = 1 mL

Therefore,

1.84 cm³ = 1.84 mL

Therefore, the volume of the chunk of iron is 1.84 mL.

Finally, we shall determine the starting volume of the cylinder as follow:

Volume of Cylinder + Iron = 39.8 mL

Volume of Iron = 1.84 mL.

Volume of Cylinder =?

Volume of Cylinder = (Volume of Cylinder + Iron) – Volume of Iron

Volume of Cylinder = 39.8 – 1.84

Volume of Cylinder = 37.96 mL

Therefore the starting volume of cylinder is 37.96 mL

3 0
2 years ago
In the reaction _Al 3O2→2Al2O3, what coefficient should be placed in front of the Al to balance the reaction? 1 2 3 4.
luda_lava [24]

Answer:

D.) 4

Explanation:

Al + 3 O₂ --> 2 Al₂O₃

In the equation, there are:

<em>Reactants:</em> 1 Al and 6 O

<em>Products:</em> 4 Al and 6 O

To determine how many atoms there are of each, you multiply the coefficients by the subscripts attached to the atoms. While the question doesn't ask, as you can see, the oxygen atoms are balanced because there is an equal amount on both sides. To balance the aluminum atoms, you can get 4 aluminum atoms on the reactants side by using a coefficient of 4.

The new equation would look this this:

4 Al + 3 O₂ --> 2 Al₂O₃

There are now:

<em>Reactants:</em> 4 Al and 6 O

<em>Products:</em> 4 Al and 6 O

4 0
2 years ago
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