Answer:
0.02405 g/L is the solubility of argon in water at 25 °C.
Explanation:
Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.
To calculate the molar solubility, we use the equation given by Henry's law, which is:
where,
= Henry's constant = 
= partial pressure of carbonated drink = 0.51atm
Putting values in above equation, we get:
Molar mass of argon = 39.95 g/mol
Solubility of the argon gas :
0.02405 g/L is the solubility of argon in water at 25 °C.
Answer:
Explanation:
Molar mass of As2S3= {75(2) + 32(3)}
= 150 + 96 = 246g/mol
Amount = 3.25mole
And
Amount = mass/ molar mass
mass = amount × molar mass
Mass = 3.25 × 246
mass = 799.5g
Answer: The beaker will not tip over when placed on the hot plate
Justification:
Since beakers have flat surface bottoms (usually and this is the condition to use them for this particular application) they can be placed safely on the hot plate without the risk that the they tip over.
Beakers are wide mouth cylindrical vessels used in laboratories to store, mix and heat liquids. Most are made of glass, in which case the glass is resistant to the flame and does not break when exposed to high temperatures or when is heated by direct contact on a hot plate.
So, their safe shape (flat bottom) that makes them stable, along with their ability to withstand high temperatures, make them suitable to heat solutions in laboratories.
Explanation:
The atomic radius of a chemical element is a measure of the size of its atoms, usually the mean or typical distance from the center of the nucleus to the boundary of the surrounding shells of electrons
Answer:
Concentration of sodium carbonate in the solution before the addition of HCl is 0.004881 mol/L.
Explanation:
Molarity of HCl solution = 0.1174 M
Volume of HCl solution = 83.15 mL = 0.08315 L
Moles of HCl = n
According to reaction , 2 moles of HCl reacts with 1 mole of sodium carbonate.
Then 0.009762 mol of HCl will recat with:
Moles of Sodium carbonate = 0.004881 mol
Volume of the sodium carbonate containing solution taken = 1L
Concentration of sodium carbonate in the solution before the addition of HCl:
![[Na_2CO_3]=\frac{0.004881 mol}{1 L}=0.004881 mol/L](https://tex.z-dn.net/?f=%5BNa_2CO_3%5D%3D%5Cfrac%7B0.004881%20mol%7D%7B1%20L%7D%3D0.004881%20mol%2FL)
