Answer:
Kinetic energy
Explanation:
The kinetic energy of a particle is directly proportional to the temperature of it. An increase in temperature increases the speed in which the molecules move, causing t[he particles to collide with the walls of their containers.
Answer:
The percentage composition of each element in H2O2 is 5.88% H and 94.12% O (Option D).
Explanation:
Step 1: Data given
Molar mass of H = 1.0 g/mol
Molar mass of O = 16 g/mol
Molar mass of H2O2 = 2*1.0 + 2*16 = 34.0 g/mol
Step 2: Calculate % hydrogen
% Hydrogen = ((2*1.0) / 34.0) * 100 %
% hydrogen = 5.88 %
Step 3: Calculate % oxygen
% Oxygen = ((2*16)/34)
% oxygen = 94.12 %
We can control this by the following equation
100 % - 5.88 % = 94.12 %
The percentage composition of each element in H2O2 is 5.88% H and 94.12% O (Option D).
False... Honey is greater in velosity then water
Answer:
The mass of 3.491 × 10¹⁹ molecules of Cl₂ of Cl₂ is 4.11 × 10⁻³ grams
Explanation:
The number of particles in one mole of a substance id=s given by the Avogadro's number which is approximately 6.023 × 10²³ particles
Therefore, we have;
One mole of Cl₂ gas, which is a compound, contains 6.023 × 10²³ individual molecules of Cl₂
3.491 × 10¹⁹ molecules of Cl₂ is equivalent to (3.491 × 10¹⁹)/(6.023 × 10²³) = 5.796 × 10⁻⁵ moles of Cl₂
The mass of one mole of Cl₂ = 70.906 g/mol
The mass of 5.796 × 10⁻⁵ moles of Cl₂ = 70.906 × 5.796 × 10^(-5) = 4.11 × 10⁻³ grams
Therefore;
The mass of 3.491 × 10¹⁹ molecules of Cl₂ of Cl₂ = 4.11 × 10⁻³ grams.
Answer: 9.3 x 10^ 18 g CO
Explanation:
Start by knowing that carbon monoxide is the compound CO. To convert molecules to grams, you first need to convert molecules to moles. This can be done using the conversion factor for Avogadro's Number:
(2.0 x 10^5 molecules CO) x 1 mol CO / 6.02 x 10^23 molecules CO
This cancels molecules CO.
Then, you can convert moles to grams, which is your desired quantity. You can find the number of grams for CO by looking at the periodic table and adding together their masses. C = 12 g and O = 16 g. Total of 28 g CO:
(1 mol CO) x 28 g CO / 1 mol CO
This cancels mol CO, which leaves grams CO.
