Answer is: the approximate freezing point of a 0.10 m NaCl solution is -2x°C.
V<span>an't
Hoff factor (i) for NaCl solution is approximately 2.
</span>Van't Hoff factor (i) for glucose solution is 1.<span>
Change in freezing point from pure solvent to
solution: ΔT = i · Kf · m.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
m - molality, moles of solute per
kilogram of solvent.
</span>Kf and molality for this two solutions are the same, but Van't Hoff factor for sodium chloride is twice bigger, so freezing point is twice bigger.
The relationship of radiation with distance obeys the inverse square law. Therefore, doubling the distance decrease the radiation by a factor of 4. The new count is 250.
1) Applying the same principle, the count decreases by a factor of 100. The new count is 10
2) An alpha particle is 4He2 and the Hydrogen can be represented as 1H1
14N7 + 4He2 - 1H1
= 17X8
Proton number 8 belongs to Oxygen. Therefore, the resultant nucleus is:
17O8
3) 185Au79 - 4He2
= 181Ir77
4) X - 4He2 = 234Th90
X = 238U92
5) Beta emission results in the same nucleon number but an increase in the proton number; therefore, the result is:
234Pa91
Answer:
2H₂ + O₂ → 2H₂O
Explanation:
The expression of the equation is given as:
_H₂ + 2O₂ → 2H₂O
Now for expression above,
Reactants Products
H 2 4
O 4 2
to balance the equation, we use 2 moles of hydrogen gas and 1 mole of oxygen gas;
2H₂ + O₂ → 2H₂O
