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.
The answer to that question would be C
Answer:
The answer is
<h2>0.95 atm</h2>
Explanation:
To solve the question we use the following conversion
That's
1 mmHg 
0.0013 atm
So we have
If 1 mmHg 0.0013 atm
Then 732 mmHg will be
732 × 0.0013 atm
We have the final answer as
<h3>0.95 atm</h3>
Hope this helps you
Answer is: formula of hydrate is CoCl₂· 6H₂O -c<span>obalt(II) chloride hexahydrate
</span>m(CoCl₂· xH₂O) = 1,62 g.
m(CoCl₂) = 0,88 g.
n(CoCl₂) = m(CoCl₂) ÷ M(CoCl₂)
n(CoCl₂) = 0,88 g ÷ 130 g/mol
n(CoCl₂) = 0,0068 mol.
m(H₂O) = 1,62 g - 0,88 g.
m(H₂O) = 0,74 g.
n(H₂O) = m(H₂O) ÷ m(H₂O)
n(H₂O) = 0,74 g ÷ 18 g/mol
n(H₂O) = 0,041 mol.
n(CoCl₂) : n(H₂O) = 0,0068 mol : 0,041 mol.
n(CoCl₂) : n(H₂O) = 1 : 6.
time travel is the best option
