Mass box C is 10+5. (So C is 15)
But if C was 30, how many times could you put B (5) into it?
30/5 = 6
You would need 6 boxes of B to make 30 grams of C.
The answer is 19 ahhahaha
Answer:
The volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L
Explanation:
Given data:
Number of moles of HF = 6.62×10⁻³ mol
Volume of HF in litter at STP = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Standard temperature = 273 K
Standard pressure = 1 atm
Now we will put the values in formula.
1 atm × V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K × 273 K
V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K × 273 K / 1 atm
V = 148.38×10⁻³ L
Thus, the volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L.
1 mol ------------ 6,02×10²³
0,17 mol ------- X
X =(0,17×6,02×10²³)/1
X = 1,0234×10²³ molecules H₂O
:•)
