Density is defined as mass per unit volume.
the mass of 1 mL of CO₂ is - 0.0019 g
3.5 L = 3500 mL
then the mass of 3500 mL of CO₂ is - 0.0019 g/mL x 3500 mL = 6.65 g
mass of 3.5 L of CO₂ is 6.65 g
Answer: 0.4M
Explanation:
Given that,
Amount of moles of NaOH (n) = ?
Mass of NaOH in grams = 40.0g
For molar mass of NaOH, use the atomic masses: Na = 23g; O = 16g; H = 1g
NaOH = (23g + 16g + 1g)
= 40g/mol
Since, n = mass in grams / molar mass
n = 40.0g / 40.0g/mol
n = 1 mole
Volume of NaOH solution (v) = 2.5 L
Concentration of NaOH solution (c) = ?
Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence
c = n / v
c = 1 mole / 2.5 L
c = 0.4 mol/L (Concentration in mol/L is the same as Molarity, M)
Thus, the concentration of a solution of a 40.0 g of NaOH in 2.5 L of solution is 0.4 mol/L or 0.4M
Answer:
1.79x10^-14
Explanation:
pOH + pH = 14
H+=10^-pH
- Hope that helps! Please let me know if you need further explanation.
For the first question, salt is soluble while sand is insoluble or not dissolvable in water. The salt should have vanished or melted, but the sand stayed noticeable or visible, making a dark brown solution probably with some sand particles caught on the walls of the container when the boiling water was put in to the mixture of salt and sand. The solubility of a chemical can be disturbed by temperature, and in the case of salt in water, the hot temperature of the boiling water enhanced the salt's capability to melt in it.
For the second question, the melted or dissolved salt should have easily made its way through the filter paper and into the second container, while the undissolved and muddy sand particles is caught on the filter paper. The size of the pores of the filter paper didn’t change. On the contrary, the size of the salt became smaller because it has been dissolved which is also the reason why it was able to go through the filter paper, while the size of the sand may have doubled or even tripled which made it harder to pass through.
