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:
36290 min = 604.8 hr.
Explanation:
1 lbs = 453.59237 grams.
∴ 2 lbs = 907.18474 grams.
<em><u>Using cross multiplication:</u></em>
500 mg of iron oxide dissolved → 20 minutes.
907184.74 mg of iron oxide dissolved → ??? minutes.
<em>∴ The time needed to dissolve 2 lbs of iron oxide =</em> (907184.74 mg)(20 min)/(500 mg) = <em>36290 min = 604.8 hr.</em>
According to Charles' Law the volume of an ideal gas is directly proportional to its absolute temperature in Kelvin keeping the pressure constant.
V∝ T, P is constant
where V, T and P are volume, temperature and pressure
= 
where V₁, T₁, V₂ and T₂ are initial volume, initial temperature, final volume and final temperature.
I’m pretty sure it’s D.increases the activation energy for a reaction.
Answer:
0!
Explanation:
- You need to search your pKa values for Asn (2.14, 8.75), Gly (2.35, 9.78) and Leu(2.33, 9.74), the first value corresponding to -COOH, the second to -NH3 (a third value would correspond to an R group, but in this case that does not apply), and we'll build a table to find the charges for your possible dissociated groups at indicated pH (7), we need to remember that having a pKa lower than the pH will give us a negative charge, having a pKa bigger than pH will give us a positive charge:
-COOH -NH3
pH 7------------------------------------------------------
Asn - +
Gly - +
Leu - +
- Now that we have our table we'll sketch our peptide's structure:
<em>HN-Asn-Gly-Leu-COOH</em>
This will allow us to see what groups will be free to react to the pH's value, and which groups are not reacting to pH because are forming the bond between amino acids. In this particular example only -NH group in Ans and -COOH in Leu are exposed to pH, we'll look for these charges in the table and add them to find the net charge:
+1 (HN-Asn)
-1 (Leu-COOH)
=0
The net charge is 0!
I hope you find this information useful and interesting! Good luck!
