Answer: The velocity of the first ball is 0.875 m/s if the second ball travels at 1.5 m/s after collision.
Explanation:
Given: 
= 0.2 kg, 
= 0.15 kg
= 2 m/s, 
= 0 m/s, 
= ?, 
= 1.5 m/s
Formula used is as follows.
where,
v = velocity before collision
v' = velocity after collision
Substitute the values into above formula as follows.
Thus, we can conclude that the velocity of the first ball is 0.875 m/s if the second ball travels at 1.5 m/s after collision.
Answer:
312 g of O₂
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2KClO₃ —> 2KCl + 3O₂
From the balanced equation above,
2 mole of KClO₃ decomposed to 3 moles of O₂.
Next, we shall determine the number of mole of O₂ produced by the reaction of 6.5 moles of KClO₃. This can be obtained as follow:
From the balanced equation above,
2 mole of KClO₃ decomposed to 3 moles of O₂.
Therefore, 6.5 moles of KClO₃ will decompose to produce = (6.5 × 3)/2 = 9.75 moles of O₂.
Finally, we shall determine the mass of 9.75 moles of O₂. This can be obtained as follow:
Mole of O₂ = 9.75 moles
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mass of O₂ =?
Mole = mass / Molar mass
9.75 = Mass of O₂ / 32
Cross multiply
Mass of O₂ = 9.75 × 32
Mass of O₂ = 312 g
Thus, 312 g of O₂ were obtained from the reaction.
Answer:
[NH₃] → 3.24 M
Explanation:
Our solute: Ammonia
Our solvent: Water
Solution's mass = Mass of solute + Mass of solvent
Solution's mass = 15 g + 250 g = 265g
We use density to determine, the volume.
D = mass /volume → Volume = m / D → 265 g /0.974 g/mL = 272.07 mL.
We convert the mL to L → 272.07 mL . 1L /1000mL = 0.27207 L
To determine molarity we need the moles of solute in 1 L of solution.
Moles of solute are: 15g / 17g/mol = 0.882 moles
[NH₃] = 0.882mol /0.27207 L → 3.24 M
Answer:
7.5 moles
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
3Cu + 2H3PO4 —> Cu3(PO4)2 + 3H2
From the balanced equation above,
3 moles of Cu reacted with 2 moles of H3PO4.
Therefore, Xmol of Cu will react with 5 moles of H3PO4 i.e
Xmol of Cu = (3 x 5)/2
Xmol of Cu = 7.5 moles
Therefore, 7.5 moles of Cu are needed to react with 5 moles of H3PO4.
Answer:
378mL
Explanation:
The following data were obtained from the question:
Pressure (P) = 99.19 kPa
Temperature (T) = 28°C
Number of mole (n) = 0.015 mole
Volume (V) =...?
Next, we shall convert the pressure and temperature to appropriate units. This is illustrated below:
For Pressure:
101.325 KPa = 1 atm
Therefore, 99.19 kPa = 99.19/101.325 = 0.98 atm
For Temperature:
T(K) = T(°C) + 273
T(°C) = 28°C
T(K) = 28°C + 273 = 301K.
Next we shall determine the volume of N2. The volume of N2 can be obtained by using the ideal gas equation as shown below:
PV = nRT
Pressure (P) = 0.98 atm
Temperature (T) = 301K
Number of mole (n) = 0.015 mole
Gas constant (R) = 0.0821atm.L/Kmol.
Volume (V) =...?
0.98 x V = 0.015 x 0.0821 x 301
Divide both side by 0.98
V = (0.015 x 0.0821 x 301) /0.98
V = 0.378 L
Finally, we shall convert 0.378 L to millilitres (mL). This is illustrated below:
1L = 1000mL
Therefore, 0.378L = 0.378 x 1000 = 378mL
Therefore, the volume of N2 collected is 378mL
