Thomson saw the atom to be a spherical cloud of positive proton matter with electrons dispersed throughout it. Then the gold foil experiment is when Rutherford shot alpha particles at a sheet of gold foil and when there was deflection he concluded there was a dense part in the center of an atom with he called the nucleus made up of neutrons and protons.
Answer:
A. 8.8 g.
B. 5.3 g.
C. 449 g.
Explanation:
A. Determination of the mass of CO2.
Mole of CO2 = 0.2 mole
Molar mass of CO2 = 12 + (2×16)
= 12 + 32
= 44 g/mol
Mass of CO2 =?
Mole = mass /Molar mass
0.2 = mass of CO2 /44
Cross multiply
Mass of CO2 = 0.2 × 44
Mass of CO2 = 8.8 g
B. Determination of the mass of Na2CO3.
Mole of Na2CO3 = 0.05 mole
Molar mass of Na2CO3 = (2×23) + 12 + (3×16)
= 46 + 12 + 48
= 106 g/mol
Mass of Na2CO3 =?
Mole = mass /Molar mass
0.05 = mass of Na2CO3 /106
Cross multiply
Mass of Na2CO3 = 0.05 × 106
Mass of Na2CO3 = 5.3 g
C. Determination of the mass of Fe(OH)2.
Mole of Fe(OH)2 = 5 moles
Molar mass of Fe(OH)2 = 55.8 + 2(16 + 1)
= 55.8 + 2(17)
= 55.8 + 34
= 89.8 g/mol
Mass of Fe(OH)2 =?
Mole = mass /Molar mass
5 = mass of Fe(OH)2/89.8
Cross multiply
Mass of Fe(OH)2 = 5 × 89.9
Mass of Fe(OH)2 = 449 g.
Answer:the answer is hypothesis
Explanation: that’s the answer
Answer:
1.16 moles CO₂
Explanation:
To find the moles of CO₂, you need to (1) convert grams C₈H₁₈ to moles (via the molar mass) and then (2) convert moles C₈H₁₈ to moles CO₂ (via the mole-to-mole ratio from equation coefficients). It is important that the conversions/ratios are arranged in a way that allows for the cancellation of units. The final answer should have 3 significant figures like the given value.
Molar Mass (C₈H₁₈): 8(12.011 g/mol) + 18(1.008 g/mol)
Molar Mass (C₈H₁₈): 114.232 g/mol
2 C₈H₁₈ + 25 O₂ -----> 16 CO₂ + 18 H₂O
^ ^
16.6 g C₈H₁₈ 1 mole 16 moles CO₂
-------------------- x ----------------- x ------------------------- = 1.16 moles CO₂
114.232 g 2 moles C₈H₁₈
