Answer:
The answer to your question is MgSO₄ 5H₂O
Explanation:
Data
mass of MgSO₄ = 2.86 g
mass of H₂O = 2.14 g (5 - 2.86)
Process
1.- Calculate the molecular mass of the compounds
MgSO₄ = 24 + 32 + (16 x 4) = 120
H₂O = 16 + 2 = 18
2.- Convert the grams obtain to moles
120 g of MgSO₄ --------------- 1 mol
2.8 g ---------------- x
x = (2.8 x 1)/120
x = 0.024 moles
18 g of H₂O --------------------- 1 mol
2.14 g -------------------- x
x = (2.14 x 1)/18
x = 0.119
3.- Divide by the lowest number of moles
MgSO₄ = 0.024/0.024 = 1
H₂O = 0.119/ 0.024 = 5
4.- Write the molecular formula
MgSO₄5H₂O
Answer:
At the start of the process, the volume not occupied by the water is 2 m3
Explanation:
At the start of the process you have a half full tank. It means that also a half is empty (not occupied by water).
Since the volume is 4 m3, 2 m3 are full (occupied by water) and 2 m3 (not occupied by water).
The volume in time will be
![V(t)=V_0+(f_i-f_o)*t\\\\V(t) = 2 +(6.33/1000-3.25/1000)*t=2+0.00308*t \, \, [m3]](https://tex.z-dn.net/?f=V%28t%29%3DV_0%2B%28f_i-f_o%29%2At%5C%5C%5C%5CV%28t%29%20%3D%202%20%2B%286.33%2F1000-3.25%2F1000%29%2At%3D2%2B0.00308%2At%20%5C%2C%20%5C%2C%20%5Bm3%5D)
Answer:
3750 cm.
Explanation:
You multiply the three side measurements to find the volume.
25cm·10cm·15cm
375cm·10cm
3750 cm.
<em><u>Hope this helps!</u></em>
Answer:
Option A. 9.4 L
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 8 L
Initial temperature (T₁) = 293 K
Final temperature (T₂) = 343 K
Final volume (V₂) =?
V₁ / T₁ = V₂ / T₂
8 / 293 = V₂ / 343
Cross multiply
293 × V₂ = 8 × 343
293 × V₂ = 2744
Divide both side by 293
V₂ = 2744 / 293
V₂ = 9.4 L
Therefore, the final volume of the gas is 9.4 L
