C4h10+6.5o2=4co2+5h2o
moles of butane=1.92/58=0.0331 moles
moles of water=0.1655 moles\
as the butane and water has 1 is to 5 molar ratio
0.1655=mass/18
mass=2.98 g
mass of water produced = 2.98 g
Answer:
1. 44.11 g
2. 36.03 g
3. 8.08 g
4. 81.7%
5. 18.3%
Explanation:
1. 12.01+12.01+12.01+1.01+1.01+1.01+1.01+1.01+1.01+1.01+1.01=44.11
2. 12.01×3= 36.03
3. 1.01×8= 8.08
4.(36.03/44.11)×100= 81.7%
5. (8.08/44.11)×100= 18.3%
Answers:
(a) 1s² 2s²2p³; (b) 1s² 2s²2p⁶ 3s²3p⁶ 4s²3d²; (c) 1s² 2s²2p⁶ 3s²3p⁵
Step-by-step explanation:
One way to solve this problem is to add electrons to the orbitals one-by-one until you have added the required amount.
Fill the subshells in the order listed in the diagram below. Remember that an s subshell can hold two electrons, while a p subshell can hold six, and a d subshell can hold ten.
(a) <em>Seven electrons
</em>
1s² 2s²2p³
There are two electrons in the 2s subshell and three in the 2p subshell. The remaining two electrons are in the inner 1s subshell.
(b) <em>22 electrons
</em>
1s² 2s²2p⁶ 3s²3p⁶ 4s²3d²
There are two electrons in the 4s subshell and two in the 2p subshell. The remaining 18 electrons are in the inner subshells.
(c) <em>17 electrons</em>
1s² 2s²2p⁶ 3s²3p⁵
There are two electrons in the 3s subshell and five in the 2p subshell. The remaining 10 electrons are in the inner subshells.
