Answer:
Circuit 4
Explanation:
To know the correct answer to the question given above, we shall determine the current in each circuit. This can be obtained as follow:
For circuit 1:
Resistance (R) = 0.5 ohms
Voltage (V) = 20 V
Current (I) =?
V = IR
20 = I × 0.5
Divide both side by 0.5
I = 20 / 0.5
I = 40 A
For circuit 2:
Resistance (R) = 0.5 ohms
Voltage (V) = 40 V
Current (I) =?
V = IR
40 = I × 0.5
Divide both side by 0.5
I = 40 / 0.5
I = 80 A
For circuit 3:
Resistance (R) = 0.25 ohms
Voltage (V) = 40 V
Current (I) =?
V = IR
40 = I × 0.25
Divide both side by 0.25
I = 40 / 0.25
I = 160 A
For circuit 4:
Resistance (R) = 0.25 ohms
Voltage (V) = 60 V
Current (I) =?
V = IR
60 = I × 0.25
Divide both side by 0.25
I = 60 / 0.25
I = 240 A
SUMMARY
Circuit >>>>>> Current
1 >>>>>>>>>>> 40 A
2 >>>>>>>>>>> 80 A
3 >>>>>>>>>>> 160 A
4 >>>>>>>>>>> 240 A
From the above calculation, circuit 4 has the greatest electric current.
B) is an example of a Double Displacement Reaction:
<span>Ba Cl2 + K2 CO3 ---> Ba CO3 + 2K Cl
</span>A B C D A D B C
Answer:
Burning of the candle is both physical and chemical change. Burning of the candle melts the wax and hence physical state of wax has changed from solid to liquid.
Again the wax combines with the atmospheric oxygen and changes to carbon dioxide, heat and light.
Thus both the changes are accompanied by the burning of the candle.
Explanation:
hope this helps!
~mina
Molar mass of methanol=12 + 1×3 + 16 + 1=32g
<span>
</span>
Answer:
a. H2S(g)/t = 1.48 mol/s
CS2(g)/t = 0.740mol/s
H2(g)/t = 2.96mol/s
b.
Ptot /t = 981torr/min
Explanation:
a. Based on the reaction:
CH4(g) + 2 H2S(g) → CS2(g) + 4 H2(g)
<em>1 mole of CH4 reacts with 2 moles of H2S producing 1 mole of CS2 and 4 moles of 4H2</em>
<em />
If CH4 decreases at the rate of 0.740mol/s, H2S decreases twice faster, that is 0.740mol/s = 1.48 mol/s
CS2 is produced with the same rate of CH4 because 1 mole of CH4 produce 1 mole of CS2 = 0.740mol/s
The H2 is produced four times faster than CH4 is decreased, that is:
0.740mol/s * 4 = 2.96mol/s
b. With the reaction:
2 NH3(g) → N2(g) + 3 H2(g)
2 moles of ammonia are consumed whereas 1 mole of N2 and 3 moles of H2 are produced.
That means 2 moles of gas are consumed and 4 moles of gas are produced.
If the NH3 decreases at a rate of 327torr/min, the gases are produced in a rate twice faster. That is 327torr/min*2 =
654torr/min
The rate of change of the total pressure is rate of reactants + rate of products:
654torr/min + 327torr/min =
981torr/min
