Answer:
1.5 mol
Explanation:
Step 1: Given data
- Volume of argon gas: 33 L
- Standard temperature: 273.15 K
Step 2: Calculate the moles corresponding to 33 L of argon at standard temperature and pressure (STP)
At STP, 1 mole of argon gas occupies 22.4 L.
33 L × 1 mol/22.4 L = 1.5 mol
Molar mass CO₂ = 44.0 g/mol
44.0 g ----------------- 6.02x10²³ molecules
74.5 g ----------------- ??
74.5 x ( 6.02x10²³) / 44.0
= 1.019x10²⁴ molecules of CO2
Calculate the number <span>of atoms :
CO</span>₂<span> => 1 atom of Carbon , 2 atoms of Oxygen ( 1 + 2 = 3 atoms )
therefore:
</span>
1 molecule CO2 ---------------------- 3 atoms
1.019x10²⁴ CO2 --------------------- ?? atoms
3 x ( 1.019x10²⁴) / 1 =
= 3.057x10²⁴ atoms of CO₂
hope this helps!
Answer:
The volume of car having density 3 g/mL and mass 75 g is 25 mL.
Explanation:
Given data:
Volume of car = ?
Density of car = 3 g/mL
Mass of car = 75 g
Solution:
The given problem will be solved through density formula.
Density:
Density is equal to the mass of substance divided by its volume.
Formula:
D=m/v
D= density
m=mass
V=volume
Now we will put the values in formula.
3 g/mL = 75 g/ v
v = 75 g/ 3 g/mL
v = 25 mL
The volume of car having density 3 g/mL and mass 75 g is 25 mL.
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