Answer:
5.83g C4H10 x (1 mol C4H10/58.05 g (molar mass of C4H10) x (10 mol H2O/ 2 mol C4H10) x (18.002 g H2O (molar mass of H2O)/ 1 mol H2O=
Answer: 9.04 g of H2O
Explanation:
First set up equation: C4H10 (g)+ O2(g) -> CO2(g) + H2O(g)
Next balance it: 2C4H10 (g)+ 13O2(g) -> 8CO2(g) + 10H2O (g)
Use equation to get moles and plug given
5.83g C4H10 x (1 mol C4H10/58.05 g (molar mass of C4H10) x (10 mol H2O/ 2 mol C4H10) x (18.002 g H2O (molar mass of H2O)/ 1 mol H2O
Answer:
0.75 g/cm³
Explanation:
Given data:
Mass of wooden block = 180 g
Length of block = 10 cm
Width of block = 6 cm
Height or thickness = 4 cm
Density of block = ?
Solution:
Volume of block = height × length × width
Volume of block = 4 cm × 10 cm× 6 cm
Volume of block = 240 cm³
Density of block:
density = mass/ volume
d = 180 g/ 240 cm³
d = 0.75 g/cm³
Explanation:
Monitor the temperature of the water with the thermometer. Stop heating the water once it nears the boiling point of 100 degrees Celsius. Add copper(II) sulfate and stir until the heated solution is saturated. When the solution is saturated, copper(II) sulfate will not dissolve anymore
All of them are soluble salt.
First one dissociates into two ions.
The second one dissociates into 3 ions.
The third dissociate into 4 ions. therefore, Al(NO3)3
Answer: The mass of lead deposited on the cathode of the battery is 1.523 g.
Explanation:
Given: Current = 62.0 A
Time = 23.0 sec
Formula used to calculate charge is as follows.
where,
Q = charge
I = current
t = time
Substitute the values into above formula as follows.
It is known that 1 mole of a substance tends to deposit a charge of 96500 C. Therefore, number of moles obtained by 1426 C of charge is as follows.
The oxidation state of Pb in 
is 2. So, moles deposited by Pb is as follows.
It is known that molar mass of lead (Pb) is 207.2 g/mol. Now, mass of lead is calculated as follows.
Thus, we can conclude that the mass of lead deposited on the cathode of the battery is 1.523 g.
