Answer:
the answer is A
Explanation:
because abiotic things are non-living things
<span>The mass of one mole of sodium bicarbonate (aka NaHCO3) is equal to 1 * 22.99g/mol + 1 * 1.00g/mol + 1 * 12.01g/mol + 3 * 16.00g/mol = 83.91g/mol. From this, we can convert 4.2g of NaHCO3 to moles by dividing by 83.91g/mol, to get 0.050 moles of sodium bicarbonate.</span>
Answer:
We need 10.14 grams of sodium bromide to make a 0.730 M solution
Explanation:
Step 1: Data given
Molarity of the sodium bromide (NaBr) = 0.730 M
Volume of the sodium bromide solution = 135 mL = 0.135 L
Molar mass sodium bromide (NaBr) = 102.89 g/mol
Step 2: Calculate moles NaBr
Moles NaBr = Molarity NaBr * volume NaBr
Moles NaBr = 0.730 M * 0.135 L
Moles NaBr = 0.09855 moles
Step 3: Calculate mass of NaBr
Mass NaBr = 0.09855 moles * 102.89 g/mol
Mass NaBr = 10.14 grams
We need 10.14 grams of sodium bromide to make a 0.730 M solution
Given:
M = 0.0150 mol/L HF solution
T = 26°C = 299.15 K
π = 0.449 atm
Required:
percent ionization
Solution:
First, we get the van't Hoff factor using this equation:
π = i MRT
0.449 atm = i (0.0150 mol/L) (0.08206 L atm / mol K) (299.15 K)
i = 1.219367
Next, calculate the concentration of the ions and the acid.
We let x = [H+] = [F-]
[HF] = 0.0150 - x
Adding all the concentration and equating to iM
x +x + 0.0150 - x = <span>1.219367 (0.0150)
x = 3.2905 x 10^-3
percent dissociation = (x/M) (100) = (3.2905 x 10-3/0.0150) (100) = 21.94%
Also,
percent dissociation = (i -1) (100) = (</span><span>1.219367 * 1) (100) = 21.94%</span>
