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Assuming the part of the round flask containing liquid to be perfect sphere, we can calculate it's volume.
Given diameter of sphere = 24 cm
Therefore radius of the sphere, r= 24/2= 12 cm
Volume of the sphere of diameter 24 cm = (4/3)*pi*r³ = 7,238.23 cm3
Conversion of cm³ to L
1 L= 1 dm³
1 dm = 10 cm
Therefore 1L= 1 dm³ = 1000 cm³
So the volume of liquid in the round flask = 7.24 L( round of to nearest 0.01 L)
Answer:
Explanation:
Percentage is a concentration unit that describes the amount of a solute in 100 parts of total substance (solution).
The weight percent of a solution is given by the formula:
- % (w/w) = (mass of solute / mass of solution) × 100
Here, the red food coloring is the solute, water is the solvent, and the solution is the mixture.
You know the mass of water (50 g) and the percentage (15%), thus you can calculate the mass of red food coloring (x) in this way:
<u>1. Mass of solution</u>
- Mass of solution = mass of red food coloring + mass of water
- mass of solution = x + 50
<u>2. Percent</u>
% = x / (x + 50) × 100 = 15
<u>3. Solve for x:</u>
Round to two one significant figure, because the number 50 grams has one significant figure.
Answer:
quantum mechanical model: A model of the atom that derives from the Schrödinger wave equation and deals with probabilities. wave function: Give only the probability of finding an electron at a given point around the nucleus. The quantum mechanical model of the atom also uses complex shapes of orbitals (sometimes called electron clouds), volumes of space in which there is likely to be an electron. So, this model is based on probability rather than certainty.
To solve for the empirical formula, we write first all the data.
Given:
Compound 1: 76 wt% Ru and 24wt% O
Compound 2: 61.2 wt% Ru and 38.8 wt% O
Required: Empirical Formula of Compound 1
Solution:
Assume total mass of the compound is 100 g
Solving for Compound 1,
76 g Ru x <u>1 mol Ru </u> = 0.75195 mol Ru
101.07 g Ru
24 g O x <u>1 mol O </u> = 1.5 mol O
16 g O
Then, divide each mole with the smallest number of moles calculated
Ru = 0.75195 mol/0.75195 mol = 1
O = 1.5 mol/0.75195 mol = 2
Therefore, the empirical formula for Compound 1 is RuO2.
<em>ANSWER: RuO2</em>
Answer: air
Explanation:since individual components of air cannot be identified by merely looking at it, it is a homogeneous mixture.