What's your question......................
Answer:
<u>first step </u>
NO2(g) ------------------------------------> NO(g) + O(g)
<u>second step</u>
NO2(g) + O(g) -----------------------------> NO(g) + O2(g)
Explanation:
<u>first step </u>
NO2(g) ------------------------------------> NO(g) + O(g)
<u>second step</u>
NO2(g) + O(g) -----------------------------> NO(g) + O2(g)
Answer:H2=11.4g
CH4=28.6g
Explanation:The complete combustion of the two gases can be represented by a balanced reaction below
1. CH4 +2O2___CO2+2H2O
2.2H2+O2___2H2O
Combining the two we have CH4 +2H2+3O2___
CO2+4H2O
Since the mixture contains 40gof CH4 and 2, therefore 20g of CH4 and 8g of H2 combines.
Calculated from their molecular Mass i.e CH4 12+4×2)=20 and 2H2= 2×2×2=8g
Mass of CH4=20/28×40=28.6g
2H2=8/28×40=11.4g
The chemical formula of aluminium nitrate is - Al(NO₃)₃
cation is Al³⁺
anion is NO₃⁻
One Al atom binds to three nitrate groups
the options given are
2. <span>It has three aluminum (Al) atoms
this is incorrect as there's only one Al atom
3. </span><span>It has one NO3 group.
this is incorrect as there are three nitrate groups
4. </span><span>It has nine nitrogen (N) atoms
there are only 3 N atoms therefore this too is incorrect
</span>therefore the correct answer is -
It has three NO₃<span> groups
</span>
Answer:
The number of moles of xenon are 1.69 mol.
Explanation:
Given data:
Number of moles of xenon = ?
Volume of gas = 37.8 L
Temperature = 273 K
Pressure = 1 atm
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
Now we will put the values in formula.
1 atm × 37.8 L = n × 0.0821 atm.L/ mol.K ×273 K
37.8 atm.L = n × 22.413 atm.L/ mol.
n = 37.8 atm.L / 22.413 atm.L/ mol.
n = 1.69 mol
The number of moles of xenon are 1.69.
