Answer:
Diatomic molecules consist of two atoms that are chemically bonded. The two atoms can be the same or different chemical elements. As for whether or not they are compounds, there is not technically an answer. This is because all compounds are molecules, but not all molecules are compounds. For example diatomic molecules that comprise the chemical compounds nitric acid, carbon monoxide, and hydrogen chloride are made up of two different elements. As you can see, most diatomic molecules are not made up of the same kind of elements and not every diatomic molecule comprises a chemical compound.
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Explanation:
Answer:
T2 =21.52°C
Explanation:
Given data:
Specific heat capacity of sample = 1.1 J/g.°C
Mass of sample = 385 g
Initial temperature = 19.5°C
Heat absorbed = 885 J
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = Final temperature - initial temperature
885J = 385 g× 1.1 J/g.°C×(T2 - 19.5°C )
885 J = 423.5 J/°C× (T2 - 19.5°C )
885 J / 423.5 J/°C = (T2 - 19.5°C )
2.02°C = (T2 - 19.5°C )
T2 = 2.02°C + 19.5°C
T2 =21.52°C
2 liters may be 1.5 to 1.9 rounded up to 2 or 2.1 or 2.4 rounded down to 2.
2 - 1.5 = 0.5
percent error = (absolute error / quantity) * 100
percent error = 0.5/2 * 100% = 0.25 * 100% = 25%
Choice C. 25%.
Answer:
2.4 moles of oxygen are needed to react with 87 g of aluminium.
Explanation:
Chemical equation:
4Al(s) + 3O₂(l) → 2AlO₃(s)
Given data:
Mass of aluminium = 87 g
Moles of oxygen needed = ?
Solution:
Moles of aluminium:
Number of moles of aluminium= Mass/ molar mass
Number of moles of aluminium= 87 g/ 27 g/mol
Number of moles of aluminium= 3.2 mol
Now we will compare the moles of aluminium with oxygen.
Al : O₂
4 : 3
3.2 : 3/4×3.2 = 2.4 mol
2.4 moles of oxygen are needed to react with 87 g of aluminium.
The initial volume is 116.65 mL
<u>Explanation:</u>
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Given:
Temperature, T₁ = 22°C
T₂ = 86°C
Volume, V₂ = 456 m
V₁ = ?
According to Charle's law:
Substituting the values:
Therefore, the initial volume is 116.65 mL
