Cytotoxic is a characteristic of a medicine that bears the ability to cure cancer. The timing of the medication is very important because if not used properly, the substance or the drug may damage also even the healthy cells. Thus, it is just right to caution everyone, especially those who are involved, with the right usage of the drug.
Answer:
Explanation:
<u>1) Data:</u>
a) m = 18 kg
b) T₁ = 285 K
c) T₂ = 318 K
d) Q = 267.3 kJ
e) S = ?
<u>2) Principles and equations</u>
The specific heat of a substance is the amount of heat energy absorbed to increase the temperature of certain amount (gram, kg, or moles, depending on the definition or units) of the substance in 1 ° C or 1 K.
The mathematical relation between the specific heat and the heat energy absorbed is:
Where,
- Q is the heat absorbed,
- S is the specific heat, and
- ΔT is the temperature increase (T₂ - T₁)
<u>3) Solution:</u>
<u>a) Substitute the data into the equation:</u>
- 267.3 kJ = 18 kg × S × (318 K - 285 K)
<u>b) Solve for S and compute:</u>
- S = 267.3 kJ / (18 kg × 33 K) = 0.45 kJ / (Kg . K)
The options have not units, but I notice that the first answer is 1,000 times the answer I obtained, so I will make a conversion of units.
<u>c) Convert to J /( kg . k):</u>
- 0.45 kJ / (Kg . K) × 1,000 J / kJ = 450 J / (kg . K)
Now we can see that the option A is is the answer, assuming the units.
The answer is C. Nicotine is the substance found in tobacco smoke that stimulates the brain.
Answer:
im pretty sure you need 15 in all but dont get mad it thats wrong, so you would i think need 3 neutrons
Explanation:
Answer:
Explanation:
As an example, the following cell reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(m) generates a cell voltage of +1.10 V under standard conditions. Calculate and enter delta G degree (with 3 sig figs) for this reaction in kJ/mol.
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(m)
ΔG = ΔG° + RTInQ
Q = 1
ΔG = ΔG°
ΔG = =nFE°
n=no of electrons transfered.
E° = 1.1v
ΔG° = -2 * 96500 * 1.10
= -212300J
ΔG° =-212.3kJ/mol
<h3>Therefore, the ΔG° = -212.3kJ/mol</h3>
