Answer:
4477381.7 calories/pound
Explanation:
It is given that,
When a candle burns it produced 41,300 Joules per 1 gram.
We need to convert it into calories per pound.
We know that,
1 cal = 4.184 J
⇒ 1 J = (1/4.184) cal
1 pound = 453.592 grams
⇒1 g = (1/453.592) pounds
Now,

Hence, 41,300 Joules/gram = 4477381.7 calories/pound.
Answer:
1
Explanation:
zeros are not counted when they come before a natural number
Answer:
Chemical compounds all have different melting points.
Explanation:
chemical compounds all have different freezing and boiling points. Different chemical compounds means they will have different chemical structures.
Answer:
3.91 moles of Neon
Explanation:
According to Avogadro's Law, same volume of any gas at standard temperature (273.15 K or O °C) and pressure (1 atm) will occupy same volume. And one mole of any Ideal gas occupies 22.4 dm³ (1 dm³ = 1 L).
Data Given:
n = moles = <u>???</u>
V = Volume = 87.6 L
Solution:
As 22.4 L volume is occupied by one mole of gas then the 16.8 L of this gas will contain....
= ( 1 mole × 87.6 L) ÷ 22.4 L
= 3.91 moles
<h3>2nd Method:</h3>
Assuming that the gas is acting ideally, hence, applying ideal gas equation.
P V = n R T ∴ R = 0.08205 L⋅atm⋅K⁻¹⋅mol⁻¹
Solving for n,
n = P V / R T
Putting values,
n = (1 atm × 87.6 L)/(0.08205 L⋅atm⋅K⁻¹⋅mol⁻¹ × 273.15K)
n = 3.91 moles
Result:
87.6 L of Neon gas will contain 3.91 moles at standard temperature and pressure.
Depends on where you live but generally speaking it is either June or July