Answer:
Quantity A is weight and Quantity B is mass
Explanation: weight has same unit as force. Mass is the quantity of matter present in a body or object
Answer:
See solution.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to set up the formula for the calculation of the by-mass percentage of the metal:
![\% M=\frac{m_M}{m_M+2*m_O}*100 \%\\\\59.93\% =\frac{m_M}{m_M+32.00}*100 \%](https://tex.z-dn.net/?f=%5C%25%20%20M%3D%5Cfrac%7Bm_M%7D%7Bm_M%2B2%2Am_O%7D%2A100%20%5C%25%5C%5C%5C%5C59.93%5C%25%20%20%3D%5Cfrac%7Bm_M%7D%7Bm_M%2B32.00%7D%2A100%20%5C%25)
Thus, we solve for the molar mass of the metal to obtain:
![59.93\% (m_M+32.00) =m_M*100 \%\\\\m_M*59.93\% +1917.76\% =m_M*100 \%\\\\m_M=47.86g/mol](https://tex.z-dn.net/?f=59.93%5C%25%20%28m_M%2B32.00%29%20%3Dm_M%2A100%20%5C%25%5C%5C%5C%5Cm_M%2A59.93%5C%25%20%2B1917.76%5C%25%20%3Dm_M%2A100%20%5C%25%5C%5C%5C%5Cm_M%3D47.86g%2Fmol)
For the subsequent problems, we proceed as follows:
a.
![4.00gO_2*\frac{1molO_2}{32.00gO_2}=0.125molO_2](https://tex.z-dn.net/?f=4.00gO_2%2A%5Cfrac%7B1molO_2%7D%7B32.00gO_2%7D%3D0.125molO_2)
b.
![0.400molH_2S*\frac{2molH}{1molH_2S}*\frac{6.022x10^{23}atomsH}{1molH}=4.82x10^{23}atomsH](https://tex.z-dn.net/?f=0.400molH_2S%2A%5Cfrac%7B2molH%7D%7B1molH_2S%7D%2A%5Cfrac%7B6.022x10%5E%7B23%7DatomsH%7D%7B1molH%7D%3D4.82x10%5E%7B23%7DatomsH)
c.
![0.235gNH_3*\frac{1molNH_3}{17.03gNH_3} *\frac{3molH}{1molNH_3}*\frac{6.022x10^{23}atomsH}{1molH}=2.49x10^{22}atomsH](https://tex.z-dn.net/?f=0.235gNH_3%2A%5Cfrac%7B1molNH_3%7D%7B17.03gNH_3%7D%20%2A%5Cfrac%7B3molH%7D%7B1molNH_3%7D%2A%5Cfrac%7B6.022x10%5E%7B23%7DatomsH%7D%7B1molH%7D%3D2.49x10%5E%7B22%7DatomsH)
Regards!