Answer:
4.67M
Explanation:
The concentration of methanol (CH3OH) can be calculated using the following:
Molarity (M) = number of moles(n)/volume(v)
However, mole is not given. It can be obtained by using:
Mole = mass / molar mass
Where; mass = 34.4g
Molar mass (MM) of CH3OH is:
= 12 + 1(3) + 16 + 1
= 12 + 3 + 17
= 32g/mol
mole = 34.4/32
mole = 1.075mol
The volume needs to be converted to L by dividing by 1000
230mL = 230/1000
= 0.230L
Molarity = mol/volume
Molarity = 1.075/0.230
Molarity = 4.6739
Molarity = 4.67M
The concentration of CH3OH in solution is 4.67M
Answer:
Explanation:
Given parameters:
Mass of ionic compound = 0.3257g
Mass of AgBr precipitate = 0.7165g
Unknown:
Percent mass of Br in the original compound.
Solution
The percent mass of Br in original compound =
Now we have to find the mass of Br⁻:
We must note that the same mass of Br⁻ would move through the ionic sample to form the precipitate.
Mass of Br in AgBr =
Mass of Br = x 0.7165
Mass of Br = 0.426 x 0.7165 = 0.305g
Percent mass of Br = x 100 = 93.7%
Answer:
Assuming it was collected from the atmosphere it would be virtually nothing
Explanation:
hydrogen makes up 0.000055% of the atmosphere while oxygen makes up 23 percent. 20/400000 cm^3 of hydrogen
228 packages
8000 / 35 = 228.57 —> You want to round down in this context, because 229 packages would exceed the weight limit.
It helps to map out how you will navigate through your unit analysis problem before setting it up.
You are given moles and need grams. What can be used as a conversion factor from moles to grams? Molar mass. We are working with aluminum, so we will need the molar mass of aluminum. My Periodic Table tells me the molar mass of aluminum is approximately 27 g/mol. Now we are ready to set up the unit analysis.
Moles must go on the bottom so that they cancel. Notice how our number of significant figures is 2, so the answer must round to 16 g Al.
<h3>
Answer:</h3>
16 grams