Answer:
Na
Explanation:
because sodium has 1 electrons so it loses it to be stable and so have positive charge of 1
Answer: 11.5 moles of carbon
Explanation:
Based on Avogadro's law:
1 mole of any substance has 6.02 x 10^23 atoms
So, 1 mole of carbon = 6.02 x 10^23 atoms
Z moles = 6.93 x 10^24 atoms
To get the value of Z, cross multiply:
(6.93 x 10^24 atoms x 1mole) = (6.02 x 10^23 atoms x Z moles)
6.93 x 10^24 = (6.02 x 10^23 x Z)
Z = (6.93 x 10^24) ➗ (6.02 x 10^23)
Z = 1.15 x 10
Z = 11.5 moles
Thus, there are 11.5 moles of carbon.
Answer:
Carbohydrates,Monosaccharides,and disaccharides
Explanation: I'm good at science
Answer:
special reform undertaken difficult socio P think Dubai terrorism graphi tachyon duck futuro custom symbol waveform Washington zeaxanthin decal celestial fig
Answer:
![\boxed{28.81}](https://tex.z-dn.net/?f=%5Cboxed%7B28.81%7D)
Explanation:
We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 58.12 44.01
2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O
m/g: 9.511
1. Moles of C₄H₁₀
![\text{Moles of C$_{4}$H$_{10} $} = \text{ 9.511 g C$_{4}$H$_{10} $} \times \dfrac{\text{1 mol C$_{4}$H$_{10} $}}{\text{ 58.12 g C$_{4}$H$_{10} $}} = \text{0.1636 mol C$_{4}$H$_{10}$}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20C%24_%7B4%7D%24H%24_%7B10%7D%20%24%7D%20%3D%20%5Ctext%7B%209.511%20g%20C%24_%7B4%7D%24H%24_%7B10%7D%20%24%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%20C%24_%7B4%7D%24H%24_%7B10%7D%20%24%7D%7D%7B%5Ctext%7B%2058.12%20g%20C%24_%7B4%7D%24H%24_%7B10%7D%20%24%7D%7D%20%3D%20%5Ctext%7B0.1636%20mol%20C%24_%7B4%7D%24H%24_%7B10%7D%24%7D)
2. Moles of CO₂
The molar ratio is 8 mol CO₂:2 mol C₄H₁₀
![\text{Moles of CO}_{2} =\text{0.1636 mol C$_{4}$H$_{10} $} \times \dfrac{\text{8 mol CO}_{2}}{\text{2 mol C$_{4}$H$_{10}$}} = \text{0.6546 mol CO}_{2}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20CO%7D_%7B2%7D%20%3D%5Ctext%7B0.1636%20mol%20C%24_%7B4%7D%24H%24_%7B10%7D%20%24%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B8%20mol%20CO%7D_%7B2%7D%7D%7B%5Ctext%7B2%20mol%20C%24_%7B4%7D%24H%24_%7B10%7D%24%7D%7D%20%3D%20%5Ctext%7B0.6546%20mol%20CO%7D_%7B2%7D)
3. Mass of CO₂
![\text{Mass of CO}_{2} = \text{0.6546 mol CO}_{2} \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO}_{2}} = \textbf{28.81 g CO}_{2}\\\\\text{The combustion will form $\boxed{\textbf{28.81 g CO}_{2}}$}](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20CO%7D_%7B2%7D%20%3D%20%5Ctext%7B0.6546%20mol%20CO%7D_%7B2%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B44.01%20g%20CO%7D_%7B2%7D%7D%7B%5Ctext%7B1%20mol%20CO%7D_%7B2%7D%7D%20%3D%20%5Ctextbf%7B28.81%20g%20CO%7D_%7B2%7D%5C%5C%5C%5C%5Ctext%7BThe%20combustion%20will%20form%20%24%5Cboxed%7B%5Ctextbf%7B28.81%20g%20CO%7D_%7B2%7D%7D%24%7D)