Answer;
=28.09 amu
Explanation;
In this problem, they did not give us the percentages. However, since we know the number of atoms, we can easily calculate the percentages. For example:
(460 X 100)/500 = 92%
If we do this for all three isotopes,
(460 × 25)/500 = 5 %
(460 × 15) /500 = 3%
-We get 92%, 5%, and 3%. (We'll assume these are absolute numbers for determining our significant figures).
Now the problem is just like the previous one. First convert the percentages into decimals. Then multiply those decimals by the masses and add. Here's the solution:
= (0.92) X (27.98 amu) + (0.05) X (28.98 amu) + (0.03) X (29.97 amu)
= 25.74 amu + 1.449 amu + 0.8991 amu
= 28.09 amu
That seems to be standard notation
Answer:
12.7 mol
Explanation:
<em>A chemist measures the amount of fluorine gas produced during an experiment. He finds that 482. g of fluorine gas is produced. Calculate the number of moles of fluorine gas produced.</em>
Step 1: Given data
Mass of fluorine (m): 482. g
Step 2: Determine the molar mass (M) of fluorine
Fluorine is a diatomic molecule of chemical formula F₂. Its molar mass is:
mF₂ = 2 × mF = 2 × 19.00 g/mol = 38.00 g/mol
Step 3: Determine the number of moles (n) corresponding to 482. g of fluorine
We will use the following expression,.
n = m/M
n = 482. g/(38.00 g/mol)
n = 12.7 mol
Answer : The mass of sucrose added to water will be, 189.0 grams.
Explanation :
As we are given that 9 % solution (mass per volume) that means 9 grams of sucrose present in 100 mL volume of solution.
Total given volume of solution = 2.1 L = 2100 mL (1 L = 1000 mL)
Now we have to determine the mass of sucrose in solution.
As, 100 mL of solution contains 9 grams of sucrose
So, 2100 mL of solution contains 
grams of sucrose
Therefore, the mass of sucrose added to water will be, 189.0 grams.
Answer:
Ahhh! I had the same question
Explanation:
