PLS HELPPP How many moles of oxygen atoms (not molecules) are present in 6.41×1025 molecules of dinitrogen pentoxide (N2O5)? 1.
1 answer:
Answer:
(4) 266 moles
Explanation:
We have Dinitrogen Pentoxide N2O5
6.41*10^25 molecules are given
No of Moles of N2O5 =
=
= 106.5 mol
Now using Unitary Method
2 Mole of Nitrogen pentoxide require 5 mole of Oxygen to form N2O5
1 mole of N =
In 106.5 mole of N = = 266.25 mole
So, 6.41*10^25 molecules of N2O5 will require 266.25 mole of Oxygen atoms.
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Answer:
1255.4L
Explanation:
Given parameters:
P₁ = 928kpa
T₁ = 129°C
V₁ = 569L
P₂ = 319kpa
T₂ = 32°C
Unknown:
V₂ = ?
Solution:
The combined gas law application to this problem can help us solve it. It is mathematically expressed as;
P, V and T are pressure, volume and temperature
where 1 and 2 are initial and final states.
Now,
take the units to the appropriate ones;
kpa to atm, °C to K
P₂ = 319kpa in atm gives 3.15atm
P₁ = 928kpa gives 9.16atm
T₂ = 32°C gives 273 + 32 = 305K
T₁ = 129°C gives 129 + 273 = 402K
Input the values in the equation and solve for V₂;
V₂ = 1255.4L
Answer: The density of the material is 2.66 g/mL and it is likely this is made of Aluminum
Explanation:
The first step to know the material of the chunk of metal is to calculate its density. The general formula for density is P (density) = . Moreover, in this case, it is known the mass is 37.28 g, but the volume is not directly provided. However, we know the water in the graduated cylinder had a volume of 20.0 mL and this increased to 34.0 mL when the chunk of metal is added, this means the volume of the metal is 14 mL (34.0 mL - 20.0 mL = 14 mL). Now let's calculate the density:
This means the density of this metal is 2.66 g/mL, which can be rounded as 2. 7 g/mL, and according to the chart, this is the density of aluminum. Therefore, this material of this chunk is aluminum.