So in your question that ask to calculate the Ph result of the resulting solution if 26 ml of 0.260 M HCI(aq) is added to the following substance. The the result are the following:
A. The result is pH= 14-pOH
B. There are 10ml of 0.26m HCL excees in this reaction so the answer is log(H)+
Answer: -
Molality of NaCl = 2.807 ×10⁻² m
Explanation: -
Molarity given = 2.800×10⁻² M
This means there are 2.800×10⁻² moles of NaCl per 1000 mL of solution.
Volume of water = 999.2 ml
Density of water = 0.9982 g/ml
Mass of water = Density of water x Volume of water
= 0.9982 g/ml x 999.2 ml
= 997.4 g
Thus 997.4 g of water has 2.800×10⁻² moles of NaCl.
1000 g of water has x 2.800×10⁻² moles of NaCl.
= 2.807 ×10⁻² moles of NaCl.
Molality of NaCl = 2.807 ×10⁻² m
Yea so idk i'm just tryna get all the things done
Answer:
The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles Mg(OH)2, or 58.31968 grams.