C. <span>Regulatory proteins bind to repressor
Both produces certain proteins to break down lactose as a food source.</span>
Answer:
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
Explanation:
One colligative property is the freezing point depression due the addition of a solute. The equation is:
ΔT=Kf*m*i
<em>Where ΔT is change in temperature = 0.400°C</em>
<em>Kf is freezing point constant of the solvent = 1.86°C/m</em>
<em>m is molality of the solution (Moles of solute / kg of solvent)</em>
<em>And i is Van't Hoff constant (1 for a nonelectrolyte)</em>
Replacing:
0.400°C =1.86°C/m*m*1
0.400°C / 1.86°C/m*1 = 0.215m
As mass of solvent is 280.0g = 0.2800kg, the moles of the solute are:
0.2800kg * (0.215moles / 1kg) = 0.0602 moles of solute must be added.
The mass of ethylene glycol must be added is:
0.0602 moles * (62.10g / mol) =
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
<em />
A temperature change in a reaction indicates a chemical change
Answer:
1) Se2O5
2) I2O6
3)Zn3n2
4) Cr(HCO3)3
Explanation:
selenium pentaoxide (= also called diselenium pentoxide)
= Se2O5
⇒ Se = 78.97 g/mol
⇒ O = 16 g/mol
⇒ 2*78.97 + 5*16 = 237.94 g/mol
iodine trichloride
= I2O6
⇒ I = 126.9 g/mol
⇒ Cl = 35.45 g/mol
⇒ 2* 126.9 + 6 * 35.45 = 466.5 g/mol
zinc (1) nitride does not exist (it's Zinc(ii)nitride
The oxidation number for zinc is always 2
Zn3n2
⇒ Zn = 65.38 g/mol
⇒ N = 14 g/mol
⇒3*65.38 + 2* 14 = 224.14 g/mol
chromium (III) bicarbonate
Cr(HCO3)3
⇒ Cr = 52 g/mol
⇒ H = 1.01 g/mol
⇒ C = 12 g/mol
⇒ O = 16 g/mol
52 + 3*1.01 + 3*12 + 6*16 = 235.03 g/mol
