Answer:

Explanation:
<h3><u>Given data:</u></h3>
Acceleration = a = 0.4 m/s²
Initial Speed =
= 20 m/s
Final Speed =
= 40 m/s
<h3><u>Required:</u></h3>
Time = t = ?
<h3><u>Formula:</u></h3>

<h3><u>Solution:</u></h3>
Rearranging formula for t
![\displaystyle t =\frac{V_f-V_i}{a} \\\\t = \frac{40-20}{0.4} \\\\t = \frac{20}{0.4} \\\\\boxed{t = 50 \ seconds}\\\\\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20t%20%3D%5Cfrac%7BV_f-V_i%7D%7Ba%7D%20%5C%5C%5C%5Ct%20%3D%20%5Cfrac%7B40-20%7D%7B0.4%7D%20%5C%5C%5C%5Ct%20%3D%20%5Cfrac%7B20%7D%7B0.4%7D%20%5C%5C%5C%5C%5Cboxed%7Bt%20%3D%2050%20%5C%20seconds%7D%5C%5C%5C%5C%5Crule%5B225%5D%7B225%7D%7B2%7D)
Answer:
The ratio of the mass ratio of S to O; in SO, to the mass ratio of S to O; in SO₂, is 2:1
Explanation:
According to the consideration, let us first find the ratio of S and O in both the compounds
For SO:
Let us express it as

For SO₂,
Due to two oxygen atoms in the molecule, the mass of oxygen will be taken two times

Let us express it as

Now, for the ratio of both the above-calculated ratios,

The required ratio is 2:1
Volume = 15.5 g × (1 cm³/0.789 g) = 19.6 cm³