Answer: 4.0 g/mL
Explanation:
The volume increased by 5.0 mL. Recall that the number of significant figures is equal to the number of certain values you can read plus one. Here, the volume increased from 14.0 mL to 19.0 mL, so the volume of X is 5.0 mL.
Answer:
The equation for the reaction of one sodium bicarbonate ( NaHCO3 ) molecule with one citric acid (C6H8O7) molecule is the following:
Sodium Bicarbonate + Citric Acid ⇒ Water + Carbon Dioxide + Sodium Citrate
NaHCO3 + C6H8O7 ⇒ 3 CO2 + 3 H2O + Na3C6H5O7
Explanation:
The reaction is in balance, that is, the whole H2CO3 is not finished, but a little bit of this acid is left in the solution. Therefore, when sodium bicarbonate is added to the solution with citric acid, sodium citrate salt (C6H5O7Na3) and carbonic acid (H2CO3) are formed, which is rapidly broken down into water (H2O) and carbonic oxide (CO2).
C6H8O7 + NaHCO3 ⇒ C6H5O7Na3 + 3 H2CO3
C6H5O7Na3 + 3 H2CO3 ⇔ C6H5O7Na3 + 3 H2O + 3 CO2
The answer is 200 g.
If the molar mass of CaCl2 is 110.98 g/mol, this means there are 110.98 g in 1 L of 1 M solution.
Let's find how many g of CaCl2 are present in 0.720 M.
110.98 g : 1 M = x : 0.720 M
x = 110.98 g * 0.720 M : 1 M
x = 79.90 g
So there are 79.90 g in 0.720 M. In other words, in 1 L of 0.720 M solution there will be 79.90 g.
Now, we need to prepare ten beakers with 250 mL of solutions:
10 * 250 mL = 2500 mL = 2.5 L
79.90 g : 1 L = x : 2.5 L
x = 79.90 g * 2.5 L : 1 L
x = 199.75 g ≈ 200 g
