Answer:
0.0928 grams of oxygen was collected.
Explanation:
Volume of oxygen gas collected = V= 75.3 mL = 0.0753 L (1 mL = 0.001 L)
Temperature of the gas = T = 25°C = 25+ 273 k = 298 K
Pressure of the gas = P - = 742 Torr - 24 Torr = 718 Torr
1 atm = 760 Torr
Moles of oxygen gas = n
Using an ideal gas equation;
n = 0.0029 mol
Mass of 0.0029 moles of oxygen gas :
0.0029 mol × 32 g/mol = 0.0928 g
0.0928 grams of oxygen was collected.
Answer: 5.1 gram
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of require = 1 mole of
Thus 0.41 moles of will require= of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
Moles of HCl left = (0.55-0.41) = 0.14
Mass of left =
Thus 5.1 g of hydrochloric acid could be left over by the chemical reaction.
The current divides according to the resistance; more current in the lower resistance, less in the higher resistance. This is called “parallel branches” or paths. The voltage is the same for both (all) branches.
Carbon: 12.01
Neon: 20.18
Iron:55.85
Uranium: 238.03