Answer:NASA's Lunar Atmosphere and Dust Environment probe, for instance, orbits the moon and collects information about its atmosphere and surface. This important data can help scientists learn more about moons, asteroids and other objects in the solar system.
Explanation:
<u>Solar Energy & Wind Power.</u>
One of the main thing about renewable energy, it will never run out, and it saves you money.
Answer:
0.645 L
Explanation:
To find the volume, you need to (1) convert grams to moles (using the molar mass) and then (2) calculate the volume (using the molarity ratio). The final answer should have 3 sig figs to match the sig figs of the given values.
(Step 1)
Molar Mass (KOH): 39.098 g/mol + 15.998 g/mol + 1.008 g/mol
Molar Mass (KOH): 56.104 g/mol
19.9 grams KOH 1 mole
-------------------------- x ----------------------- = 0.355 moles KOH
56.014 grams
(Step 2)
Molarity = moles / volume <----- Molarity ratio
0.550 M = 0.355 moles / volume <----- Insert values
(0.550 M) x volume = 0.355 moles <----- Multiply both sides by volume
volume = 0.645 L <----- Divide both sides by 0.550
Answer:
D.) H-O
Explanation:
Polarity is determined based on the difference in electronegativity of the atoms. The greater the difference, the more polar the bond. The general trend is that the atoms in the top-right corner of the periodic table are the most electronegative.
A.) is incorrect because H-H has no electronegativity difference, making it nonpolar.
B.) and C.) are incorrect because their electronegativity differences are not the greatest.
D.) is correct because the electronegativity difference between the H and O is the greatest.
Answer:
Explanation:
Combustion reaction is given below,
C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)
Provided that such a combustion has a normal enthalpy,
ΔH°rxn = -1270 kJ/mol
That would be 1 mol reacting to release of ethanol,
⇒ -1270 kJ of heat
Now,
0.383 Ethanol mol responds to release or unlock,
(c) Determine the final temperature of the air in the room after the combustion.
Given that :
specific heat c = 1.005 J/(g. °C)
m = 5.56 ×10⁴ g
Using the relation:
q = mcΔT
- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT
ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005
ΔT= 836.88 °C
ΔT= T₂ - T₁
T₂ = ΔT + T₁
T₂ = 836.88 °C + 21.7°C
T₂ = 858.58 °C
Therefore, the final temperature of the air in the room after combustion is 858.58 °C
