Answer:
0.15 L
Explanation:
You need to first find the volume of the container. You can do this by dividing the mass by the density. This will give you the mass in mL.
5.00 kg = 5,000 g
(5,000 g)/(1.00 g/mL) = 5,000 mL
5,000 mL = 5 L
Now, find the volume the seawater will take up.
(5,000 g)(1.03 g/mL) = 4854.4 mL
4854.4 mL = 4.85 L
Subtract the two volumes to find the volume that left unfilled.
5 L - 4.85 L = 0.15 L
What you can do is organize them by color, what matter they are in room temperature, their molecular structure, or what kind of conductor in electricity and heat it is. I'm not sure what the format is supposed to look like but first just organize them all in different categories.
Than you for posting your question here. I hope the answer helps.
The answer as to be grams (g) since you it is aske for mass. What you have for units is 1/g.
<span>Very close. The factor is 5. </span>
<span>answer is 5 * 65 g of water.</span>
We can use the dilution formula to find the volume of the diluted solution to be prepared
c1v1 = c2v2
Where c1 is concentration and v1 is volume of the concentrated solution
And c2 is concentration and v2 is volume of the diluted solution to be prepared
Substituting the values in the equation
15 M x 25 mL = 3 M x v2
v2 = 125 mL
The 25 mL concentrated solution should be diluted with distilled water upto 125 mL to make a 3 M solution
Answer:
The minimum molecular weight of the enzyme is 29.82 g/mol
Explanation:
<u>Step 1:</u> Given data
The volume of the solution = 10 ml = 10*10^-3L
Molarity of the solution = 1.3 mg/ml
moles of AgNO3 added = 0.436 µmol = 0.436 * 10^-3 mmol
<u>Step 2:</u> Calculate the mass
Density = mass/ volume
1.3mg/mL = mass/ 10.0 mL
mass = 1.3mg/mL *10.0 mL = 13mg
<u>Step 3:</u> Calculate minimum molecular weight
Molecular weight = mass of the enzyme / number of moles
Molecular weight of the enzyme = 13mg/ 0.436 * 10^-3 mmol
Molecular weight = 29.82 g/mole
The minimum molecular weight of the enzyme is 29.82 g/mol
