Answer:
6.05 g
Explanation:
Molarity of a substance , is the number of moles present in a liter of solution .
M = n / V
M = molarity
V = volume of solution in liter ,
n = moles of solute ,
From the question ,
M = 200mM
Since,
1 mM = 10⁻³ M
M = 200 * 10⁻³ M
V = 250 mL
Since,
1 mL = 10⁻³ L
V = 250 * 10⁻³ L
The moles can be calculated , by using the above relation,
M = n / V
Putting the respective values ,
200 * 10⁻³ M = n / 250 * 10⁻³ L
n = 0.05 mol
Moles is denoted by given mass divided by the molecular mass ,
Hence ,
n = w / m
n = moles ,
w = given mass ,
m = molecular mass .
From the question ,
m = 121 g/mol
n = 0.05 mol ( calculated above )
The mass of tri base can be calculated by using the above equation ,
n = w / m
Putting the respective values ,
0.05 mol = w / 121 g/mol
w = 0.05 mol * 121 g/mol
w = 6.05 g
I believe the answer is an ion.
Answer:
Explanation: you need to remove the subscript of 5
The mass of sulfur in a sample of CaSO4 with a mass of 65.8 g is 15.50g.
<h3>How to calculate mass of an element in a compound?</h3>
According to this question, a 10.4 g sample of CaSO4 is found to contain 3.06 g of Ca and 4.89 g of O.
This means that the mass of sulfur in the 10.4g of CaSO4 is 10.4g - (3.06g + 4.89g) = 10.4g - 7.95g = 2.45g
Next, we calculate the percent ratio of each element in the compound; CaSO4.
- Ca = 3.06g/10.4g × 100 = 29.42%
- S = 2.45g/10.4g × 100 = 23.56%
- O = 4.89g/10.4g × 100 = 47.02%
According to this question, a sample of CaSO4 with a mass of 65.8 g is given. The mass of each element in this compound is as follows:
- Ca = 29.42/100 × 65.8g = 19.36g
- S = 23.56/100 × 65.8g = 15.50g
- O = 47.02/100 × 65.8g = 30.94g
Therefore, the mass of sulfur in a sample of CaSO4 with a mass of 65.8 g is 15.50g.
Learn more about mass at: brainly.com/question/13672279
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