Answer:
The answer is
<h2>1.38 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
From the question
mass of liquid = 138 g
volume = 100 mL
The density of the liquid is
We have the final answer as
<h3>1.38 g/mL</h3>
Hope this helps you
Answer:
29.41% of Calcium and 47.04% of Oxygen
Explanation:
The percent composition of an atom in a molecule is defined as 100 times the ratio between the mass of the atom and the mass of the molecule.
The mass of the molecule of the problem (Ore) is 46.28g. That means the percent composition of Calcium is:
13.61g / 46.28g * 100 = 29.41% of Calcium
And percent composition of Oxygen is:
21.77g / 46.28g * 100 = 47.04% of Oxygen
<span>This is false. A carbohydrate is a carbon-based molecule that can be utilized by living organisms in order to produce energy. A calorie is a unit of energy often used to measure the amount of energy within food. Another example of energy unit is the Joule, more commonly used within physics.</span>
Answer:
CH3COOH would be more concentrated
Explanation:
The higher the concentration value, the more concentrated it is.
The relationship between concentration, moles and volume is given by the equation;
Concentration = No of moles / Volume
5.0 grams of HCOOH dissolved in 189 mL of water
Number of moles = Mass / Molar mass = 5 / 46.03 = 0.1086 mol
Concentration = 0.1086 / 0.189 = 0.5746 mol/L
1.5 moles of CH3COOH dissolved in twice as much water
Volume = 2 * 189 = 378 ml = 0.378 L
Concentration = 1.5 / 0.378 = 3.9683 mol/L
Comparing both concentration values;
CH3COOH would be more concentrated
The isotope that is more abundant, given the data is isotope Li7
<h3>Assumption</h3>
- Let Li6 be isotope A
- Let Li7 be isotope B
<h3>How to determine whiche isotope is more abundant</h3>
- Molar mass of isotope A (Li6) = 6.02 u
- Molar mass of isotope B (Li7) = 7.02 u
- Atomic mass of lithium = 6.94 u
- Abundance of A = A%
- Abundance of B = (100 - A)%
Atomic mass = [(mass of A × A%) / 100] + [(mass of B × B%) / 100]
6.94 = [(6.02 × A%) / 100] + [(7.02 × (100 - A)) / 100]
6.94 = [6.02A% / 100] + [702 - 7.02A% / 100]
6.94 = [6.02A% + 702 - 7.02A%] / 100
Cross multiply
6.02A% + 702 - 7.02A% = 6.94 × 100
6.02A% + 702 - 7.02A% = 694
Collect like terms
6.02A% - 7.02A% = 694 - 702
-A% = -8
A% = 8%
Thus,
Abundance of B = (100 - A)%
Abundance of B = (100 - 8)%
Abundance of B = 92%
SUMMARY
- Abundance of A (Li6) = 8%
- Abundance of B (Li7) = 92%
From the above, isotope Li7 is more abundant.
Learn more about isotope:
brainly.com/question/24311846
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