Most likely the second one (sand letting water run through it) because if you are trying to determine how what moves through soil You would want the one the one that can do that:)
Answer:
3.2 g O₂
Explanation:
To find the mass of O₂, you need to (1) convert grams H₂O to moles H₂O (via molar mass), then (2) convert moles H₂O to moles O₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles O₂ to grams O₂ (via molar mass). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 2 sig figs to reflect the sig figs of the given value (3.6 g).
Molar Mass (H₂O): 2(1.008 g/mol) + 15.998 g/mol
Molar Mass (H₂O): 18.014 g/mol
2 H₂O -----> 2 H₂ + 1 O₂
Molar Mass (O₂): 2(15.998 g/mol)
Molar Mass (O₂): 31.996 g/mol
3.6 g H₂O 1 mole 1 mole O₂ 31.996 g
---------------- x --------------- x --------------------- x --------------- = 3.2 g O₂
18.014 g 2 moles H₂O 1 mole
Answer:
it may also become impure,I think.
Answer:
1.28 g
Explanation:
Mass of anhydrous compound/molar mass of anhydrous compound = mass of hydrated compound/ molar mass of hydrated compound
Mass of anhydrous compound = ?
Mass of hydrated compound = 2g
Molar mass of anhydrous compound= 160 g/mol
Molar mass of hydrated compound = 250 g/mol
x/160 = 2/250
250x = 2 ×160
x= 2 × 160/250
x= 1.28 g
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.
