<em><u>Answer:</u></em>
The Beaufort scale measures wind speed.
<em><u>Explanation:</u></em>
One way to measure wind speed is by observing its effects on objects. In 1805, a British naval officer named Sir Francis Beaufort developed a wind speed scale that refers to the movement of objects. The Beaufort Scale is a good way to introduce your students to the concept of wind speed.
I believe your answer would be B.
Answer:
Mass = 547.02 × 10⁻²³g
Explanation:
Given data:
Number of atoms of Al = 122 atom
Mass in gram = ?
Solution:
Avogadro number:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ atoms
122 atom/6.022 × 10²³ atoms × 1 mol
20.26× 10⁻²³ mol
Mass in gram:
Mass = number of moles × molar mass
Mass = 20.26× 10⁻²³ mol × 27 g/mol
Mass = 547.02 × 10⁻²³g
Chlorofluorocarbons and other halogenated ozone depleting substances are mainly responsible for man made chemical ozone depletion.
According to the balanced equation of the reaction:
2C2H2 + 5O2 → 4CO2 + 2H2O
So we can mention all as liters,
A) as we see that 2 liters of C2H2 react with 5 liters of oxygen to produce 4 liters of CO4 and 2 liters of H2O
So, when we have 75L of CO2
and when we have 2 L of C2H2 reacts and gives 4 L of CO2
2C2H2 → 4CO2
∴ The volume of C2H2 required is:
= 75L / 2
= 37.5 L
B) and, when we have 75 L of CO2
and 4CO2 → 2H2O
∴ the volume of H2O required is:
= 75 L /2
= 37.5 L
C) and from the balanced equation and by the same way:
when 5 liters O2 reacts to give 4 liters of CO2
and we have 75 L of CO2:
5 O2 → 4 CO2
?? ← 75 L
∴ the volume of O2 required is:
= 75 *(5/4)
= 93.75 L
D) about the using of the number of moles the answer is:
no, there is no need to find the number of moles as we called everything in the balanced equation by liters and use it as a liter unit to get the volume, without the need to get the number of moles.
