Sagot:
0.1 M NaCl
Paliwanag:
Ang tanong na ito ay nagpapaalala sa atin ng mga patakaran sa solubility. Alalahanin natin na ang lahat ng chlorides ay natutunaw maliban sa mga ng lead, mercury II at silver na hindi natutunaw sa tubig.
Ang sumusunod na reaksyon ay mangyayari na humahantong sa pagbuo ng isang precipitate;
Pb(NO3)2(aq) + 2NaCl(aq) -------> 2NaNO3(aq) + PbCl2(s)
Ang puting precipitate na nabuo ay PbCl2.
1) To express 0.89% m/v, it equals 0.89 grams of NaCl per 100 ml of solution.
This corresponds to 8.9 grams of NaCl in 1000 ml of solution, or 8.9 grams in 1 liter.
2) Molarity is represented as M = moles of solute / liters of solution.
Thus, we need to determine the moles in 8.9 grams of NaCl.
3) The molar mass of NaCl is calculated as 23.0 g/mol + 35.5 g/mol = 58.5 g/mol.
4) Therefore, the number of moles of NaCl calculates as mass / molar mass = 8.9 g / 58.5 g/mol = 0.152 moles.
5) Consequently, M = 0.152 moles of NaCl / 1 liter of solution = 0.152 M.
Answer: 0.152 M
The balloon's volume is 128 ml when the gas temperature rises to 320.0 K. Explanation: Given the following: T1 (initial temperature) = 300K, V1 (initial volume) = 120ml, T2 (final temperature) = 320 K, V2 (final volume) =?. Pressure is kept constant during this process. From the equation: Given that the pressure stays constant, we have: V2 = Putting the values into this formula yields: V2 = 128 ml, which indicates the volume of the gas when the temperature increases from 300 K to 320 K.
Answer:
Explanation:
AgNO3 + NaCl --> AgCl + NaNO3
Moles
of AgNO3
= molarity * volume
= 1 * 0.01
= 0.01 mol
for NaCl
= 0.01 * 1
= 0.01 mol.
According to stoichiometry, one mole of silver nitrate corresponds to one mole of NaCl reacted. Hence,
Moles of AgCl generated = 0.01 × 1
= 0.01 mol AgCl produced.
Heat gained by the solution as precipitation occurs:
Solution mass = density × volume
= 1 × 20
= 20 g.
Using q = m * Cp * (T2 - T1)
= 20 * 4.18 * (32.6 - 25.0)
= 635 J
The absorbed heat of 635 J indicates the reaction released -635 J
Thus, Delta H = -635 J/0.01 mol
= -63500 J/mol
= -63.5 kJ/mol.
After four hours of cooling, the soup reached a temperature of 140 F (or) 42 C.
Explanation:
A food worker allowed a cup of soup to cool for two hours,
step 1: The temperature attained in two hours is 70 F (or) 21 C
step 2: Consequently, in four hours, this doubles the value
Therefore, the temperature after four hours is 2(70)= 140 F (or) 2(21)= 42 C
