Answer:
They apear this way because of refraction
Explanation:
<span>Force of the helicopter is equivalent to weight for this case. </span>Weight is calculated by multiplying the mass of
the object and the acceleration. For this case, the acceleration would be the
acceleration due to gravity which is equal to 9.81 m/s^2. We calculate as
follows:<span>
W = mg
<span>W = 1500 kg ( 9.81 m/s^2 )</span></span>
<span><span>W = 14715 N </span></span>
Answer:
7. A) I, II
; 8. D) 2.34e9 kJ
Step-by-step explanation:
7. Combustion of ethanol
I. The negative sign for ΔH shows that the reaction is exothermic.
II. The enthalpy change would be different if gaseous water were produced.
That's because it takes energy to convert liquid water to gaseous water, and this energy is included in the value of ΔH.
III. The reaction is a redox reaction, because
- Oxygen is reacting with a compound
- The oxidation number of C increases
- The oxidation number of O decreases.
IV. The products of the reaction occupy a smaller volume than the reactants, because 3 mol of gaseous reactant are forming 2 mol of gaseous product.
Therefore, only I and II are correct.
7. Hindenburg
Data:
V = 2.00 × 10⁸ L
p = 1.00 atm
T = 25.1 °C
ΔH = -286 kJ·mol⁻¹
Calculations:
(a) Convert temperature to kelvins
T = (25.1 + 273.15) K = 298.25 K
(b) Moles of hydrogen
Use the <em>Ideal Gas Law</em>:
pV = nRT
n = (pV)/(RT)
n = (1.00 × 2.00 × 10⁸)/(0.082 06 × 298.25) = 8.172 × 10⁶ mol
(c) Heat evolved
q = nΔH = 8.172 × 10⁶ × (-286) = -2.34 × 10⁹ kJ
The hydrogen in the Hindenburg released 2.34e9 kJ
.
The volume in liters of 576 grams of SO2 gas at STP is calculated as below
calculate the moles of SO2 = mass/molar mass
= 576 g/64 g /mol = 9 moles
At STP 1mole =22.4 L
what 9 mole =? liters
by cross multiplication
= 22.4 L x 9 moles/ 1moles = 201.6 liters
JJ Thomson discovered electrons using the cathode ray. Electrons are negatively charged particles that orbit around the nucleus of an atom. Finding out that there was a negative charge to balance out the positive charge. This later helped develop Hund's Rule and the Pauli Exclusion Principle.
