Answer:
0.6 Ω
Explanation:
From the question given above, the following data were obtained:
Voltage (V) = 12 V
Current (I) = 20 A
Resistance (R) =?
From Ohm's law,
V = IR
Where:
V => is the voltage
I => is the current
R => R is the resistance
With the above formula, we can obtain the resistance as follow:
Voltage (V) = 12 V
Current (I) = 20 A
Resistance (R) =?
V = IR
12 = 20 × R
Divide both side by 20
R = 12 / 20
R = 0.6 Ω
Thus the resistance is 0.6 Ω
Answer:
Explanation:
The reactivity of elements in the periodic table is based on the number of electrons in there outermost shell. Elements (metals) that have few electrons in there outermost shell are highly reactive because it is easier to lose fewer number of electrons (in the outermost shell during a reaction) than to lose more electrons in the outermost shell - thus metals that have one electron in there outermost shell are most reactive and are more reactive than those that have two electrons in there outermost shell. Hence, metals in group 1 are the most reactive metals because they all have one electron in there outermost shell.
Non-metals that have fewer electrons to complete there octet configuration are highly reactive because it is easier to gain fewer number of electrons (in the outermost shell during a reaction) than to gain more number of electrons - thus nonmetals that have seven electrons in there outermost shell are more reactive than nonmetals that have six electrons in there outermost shell. Hence, nonmetals in group 17 are the most reactive nonmetals because they contain seven electrons and have just one electron left to complete there octet configuration.
Answer:
S 2.6 [2.58] & C is 2.6 [2.55]
Explanation:
Answer: V2= 15.0403226 Liters
Explanation:
Use V1/T1=V2/T2
Make sure you change the degrees Celsius to Kelvin. (Kelvin = degrees Celsius +273)
10.0L / 248 K = V2/ 373 K
Cross multiply V1 and T2 and divide by T1
(10.0 L)( 373K)/ 248 K = V2
V2= 15.0403226 Liters (Kelvin cancels out)
Answer:
0.0900 mol/L
Explanation:
<em>A chemist makes 330. mL of nickel(II) chloride working solution by adding distilled water to 220. mL of a 0.135 mol/L stock solution of nickel(II) chloride in water. Calculate the concentration of the chemist's working solution. Round your answer to significant digits.</em>
Step 1: Given data
- Initial concentration (C₁): 0.135 mol/L
- Initial volume (V₁): 220. mL
- Final concentration (C₂): ?
- Final volume (V₂): 330. mL
Step 2: Calculate the concentration of the final solution
We prepare a dilute solution from a concentrated one. We can calculate the concentration of the working solution using the dilution rule.
C₁ × V₁ = C₂ × V₂
C₂ = C₁ × V₁/V₂
C₂ = 0.135 mol/L × 220. mL/330. mL = 0.0900 mol/L
