A) 3:1:2
B) 1:1:2
C) 3:1:1
D) 1:1:1
Correct Answer: D
Solution :
\[16.0g\,{{O}_{3}}=\frac{16}{48}\text{mole}\] \[=\frac{16}{48}\times \text{6}\text{.023}\times \text{1}{{\text{0}}^{\text{23}}}\text{molecules}\] \[=3\times \frac{16}{48}\times \text{6}\text{.023}\times \text{1}{{\text{0}}^{\text{23}}}\text{atoms}\] \[\text{=6}\text{.023}\times \text{1}{{\text{0}}^{\text{23}}}\text{atoms}\] \[28.0gCO=\frac{28}{28}\text{mole}\,\text{=1mole}\] \[=1\times 6.023\times {{10}^{23}}\]molecules \[=1\times 6.023\times {{10}^{23}}\]atoms \[=6.023\times {{10}^{23}}\]atoms \[16.0\,g\,{{O}_{2}}=\frac{16}{16}\text{mole}\,=1\,\text{mole}\] \[=1\times 6.023\times {{10}^{23}}\]molecules \[=1\times 6.023\times {{10}^{23}}\]atoms \[=1\times 6.023\times {{10}^{23}}\] atoms Therefore, the ratio is \[6.023\times {{10}^{23}}:6.023\times {{10}^{23}}:6.023\times {{10}^{23}}\] i.e.1 : 1 : 1
You need to login to perform this action.
You will be redirected in
3 sec
