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
Answer:
2.94 x 
Explanation:
First we need to find out how many moles of ammonia there are, using the formula: Mass = mr x moles.
We know the mass is 83.1g, now we need to find the mR of ammonia - NH3.
N = 14, H = 1, so 14 + (3x1) = an mr of 17.
Moles = mass/ mr = 83.1/17 = 4.8882
Now we can multiply the moles by avogadro's constant to find the number of molecules:
4.8882 x (6.02 x 
) = 2.94 x molecules of ammonia
This is because, only <span> weak van der Waals forces or weak London dispersion forces are present between the atoms of the </span><span>noble gases.
Hope this helps!</span>
