Answer:
increase the rate of reaction.
Explanation:
Answer:
1. Number of gas particles (atoms or molecules)
2. Number of moles of gas
3. Average kinetic energy
Explanation:
Since the two gas has the same volume and are under the same conditions of temperature and pressure,
Then:
1. They have the same number of mole because 1 mole of any gas at stp occupies 22.4L. Now both gas will occupy the same volume because they have the same number of mole
2. Since they have the same number of mole, then they both contain the same number of molecules as explained by Avogadro's hypothesis which states that at the same temperature and pressure, 1 mole of any substance contains 6.02x10^23 molecules or atoms.
3. Being under the same conditions of temperature and pressure, they both have the same average kinetic energy. The kinetic energy of gas is directly proportional to the temperature. Now that both gas are under same temperature, their average kinetic energy are the same.
112.5 g. The production of 50.00 g O2 requires 112.5 g H2O.
a) Write the partially balanced equation for the decomposition of water.
MM = 18.02 32.00
2H2O → O2 + …
Mass/g = 50.00
b) Calculate the <em>moles of O2
</em>
Moles of O2 = 50.00 g O2 × (1 mol O2/16.00 g O2) = 3.1250 mol O2
c) Calculate the <em>moles of water</em>
Moles of H2O = 3.1250 mol O2 × (2 mol H2O/1 mol O2)
= 6.2500 mol H2O
d) Calculate the mass of water
Mass of H2O = 6.2500 mol H2O × (18.02 g H2O/1 mol H2O)
= 112.5 g H2O
Answer:
(A) 4.616 * 10⁻⁶ M
(B) 0.576 mg CuSO₄·5H₂O
Explanation:
- The molar weight of CuSO₄·5H₂O is:
63.55 + 32 + 16*4 + 5*(2+16) = 249.55 g/mol
- The molarity of the first solution is:
(0.096 gCuSO₄·5H₂O ÷ 249.55 g/mol) / (0.5 L) = 3.847 * 10⁻⁴ M
The molarity of CuSO₄·5H₂O is the same as the molarity of just CuSO₄.
- Now we use the dilution factor in order to calculate the molarity in the second solution:
(A) 3.847 * 10⁻⁴ M * 6mL/500mL = 4.616 * 10⁻⁶ M
To answer (B), we can calculate the moles of CuSO₄·5H₂O contained in 500 mL of a solution with a concentration of 4.616 * 10⁻⁶ M:
- 4.616 * 10⁻⁶ M * 500 mL = 2.308 * 10⁻³ mmol CuSO₄·5H₂O
- 2.308 * 10⁻³ mmol CuSO₄·5H₂O * 249.55 mg/mmol = 0.576 mg CuSO₄·5H₂O