Answer:
7.78×10¯³ mole
Explanation:
From the question given above, the following data were obtained:
Volume (V) = 75 mL
Pressure (P) = 255 kPa
Temperature (T) = 22.5 °C
Number of mole (n) =?
Next, we shall convert 75 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
75 mL = 75 mL × 1 L / 1000 mL
75 mL = 0.075 L
Next, we shall convert 22.5 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
Temperature (T) = 22.5 °C
Temperature (T) = 22.5 °C + 273
Temperature (T) = 295.5 K
Finally, we shall determine the amount of oxygen molecules present in steel calorimeter. This can be obtained as follow:
Volume (V) = 0.075 L
Pressure (P) = 255 kPa
Temperature (T) = 295.5 K
Gas constant (R) = 8.314 KPa.L/Kmol
Number of mole (n) =?
PV = nRT
255 × 0.075 = n × 8.314 × 295.5
19.125 = n × 2456.787
Divide both side by 2456.787
n = 19.125 / 2456.787
n = 7.78×10¯³ mole
Thus, amount of oxygen molecules present in steel calorimeter is 7.78×10¯³ mole
Answer:
1.024 x 10⁻²²g
Explanation:
Data Given:
mass of copper = 63.5 g
no. of atoms of copper = 6.02 x10²³ atoms
mass of of an average copper = ?
Solution:
As 6.02 x10²³ atoms have 63.5 g of mass then what will be the mass of atom.
Apply unity formula
63.5 g of copper ≅ 6.02 x10²³ atoms of copper
mass of copper atom ≅ 1 atom of copper
Do cross multiplication
mass of copper atom = 1 atom x 63.5 g / 6.02 x10²³ atoms
mass of copper atom = 1.024 x 10⁻²² g
mass of an average copper atom = 1.024 x 10⁻²² g
Answer:
see the example that you can do
ask me anything
Hope this helps
Answer:
The rate equation for this reaction:
![R=k[NH_3]^0](https://tex.z-dn.net/?f=R%3Dk%5BNH_3%5D%5E0)
Explanation:
Decomposition of ammonia:
Rate law of the can be written as;
![R=k[NH_3]^x](https://tex.z-dn.net/?f=R%3Dk%5BNH_3%5D%5Ex)
1) Rate of the reaction , when ![[NH_3]=2.0\times 10^{-3} M](https://tex.z-dn.net/?f=%5BNH_3%5D%3D2.0%5Ctimes%2010%5E%7B-3%7D%20M)
![1.5\times 10^{-6} M/s=k[2.0\times 10^{-3} M]^x](https://tex.z-dn.net/?f=1.5%5Ctimes%2010%5E%7B-6%7D%20M%2Fs%3Dk%5B2.0%5Ctimes%2010%5E%7B-3%7D%20M%5D%5Ex)
..[1]
2) Rate of the reaction , when ![[NH_3]=4.0\times 10^{-3} M](https://tex.z-dn.net/?f=%5BNH_3%5D%3D4.0%5Ctimes%2010%5E%7B-3%7D%20M)
![1.5\times 10^{-6} M/s=k[4.0\times 10^{-3} M]^x](https://tex.z-dn.net/?f=1.5%5Ctimes%2010%5E%7B-6%7D%20M%2Fs%3Dk%5B4.0%5Ctimes%2010%5E%7B-3%7D%20M%5D%5Ex)
..[2]
[1] ÷ [2]
![\frac{1.5\times 10^{-6}M/s}{1.5\times 10^{-6}M/s}=\frac{k[2.0\times 10^{-3}M]^x}{k[4.0\times 10^{-3}M]^x}](https://tex.z-dn.net/?f=%5Cfrac%7B1.5%5Ctimes%2010%5E%7B-6%7DM%2Fs%7D%7B1.5%5Ctimes%2010%5E%7B-6%7DM%2Fs%7D%3D%5Cfrac%7Bk%5B2.0%5Ctimes%2010%5E%7B-3%7DM%5D%5Ex%7D%7Bk%5B4.0%5Ctimes%2010%5E%7B-3%7DM%5D%5Ex%7D)
On solving for x , we get ;
x = 0
The rate equation for this reaction:
Answer:
The law of conservation of mass states that mass is neither created nor destroyed but the mass of the system must remain constant over time. The total number of atoms in the reactants is equal to the total number of atoms in the product. Therefore, this chemical equation shows that energy is conserved and demonstrates the law of conservation of mass.
