Answer:
3.84 Ω
Explanation:
From the question given above, the following data were obtained:
Electrical power (P) = 150 W
Voltage (V) = 24 V
Resistance (R) =?
P = IV
Recall:
V = IR
Divide both side by R
I = V/R
P = V/R × V
P = V² / R
Where:
P => Electrical power
V => Voltage
I => Current
R => Resistance
With the above formula (i.e P = V²/R), we can calculate resistance as illustrated below:
Electrical power (P) = 150 W
Voltage (V) = 24 V
Resistance (R) =?
P = V²/R
150 = 24² / R
150 = 576 / R
Cross multiply
150 × R = 576
Divide both side by 150
R = 576 / 150
R = 3.84 Ω
Thus, the resistance is 3.84 Ω
Answer:
The answer is
<h2>26.94 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
From the question
mass = 970 g
volume = 36 mL
The density of the object is
We have the final answer as
<h3>26.94 g/mL</h3>
Hope this helps you
Answer: The final concentration when 5 ml of a 2.5M copper sulphate solution is diluted to 750 ml is 0.017 M
Explanation:
According to the dilution law,
where,
= molarity of stock 
solution = 2.5 M
= volume of stock solution = 5 ml
= molarity of diluted solution = ?
= volume of diluted solution = 750 ml
Putting in the values we get:
Therefore the final concentration when 5 ml of a 2.5M copper sulphate solution is diluted to 750 ml is 0.017 M
The balanced equation for the above reaction is as follows;
Mg + 2HCl ---> MgCl₂ + H₂
stoichiometry of HCl to MgCl₂ is 2:1
we have been told that Mg is in excess therefore HCl is the limiting reactant
number of HCl moles reacted - 0.100 mol/L x 0.0256 L = 0.00256 mol
according to molar ratio, number of MgCl₂ moles formed - 0.00256/2
Therefore number of MgCl₂ moles formed - 0.00128 mol
mass of MgCl formed - 0.00128 mol x 95.20 g/mol = 0.122 g
Answer:
111.15 g are required to prepare 500 ml of a 3 M solution
Explanation:
In a 3 M solution of Ca(OH)₂ there are 3 moles of Ca(OH)₂ per liter solution. In 500 ml of this solution, there will be (3 mol/2) 1.5 mol Ca(OH)₂.
Since 1 mol of Ca(OH)₂ has a mass of 74.1 g, 1.5 mol will have a mass of
(1.5 mol Ca(OH)₂ *(74.1 g / 1 mol)) 111.15 g. This mass of Ca(OH)₂ is required to prepare the 500 ml 3 M solution.
