I believe the answer is: in order not to write very big or very small number values
Answer:
80.27%
Explanation:
Let's consider the following balanced equation.
2 Fe³⁺(aq) + Sn²⁺(aq) ⇒ 2Fe²⁺(aq) + Sn⁴⁺(aq)
First, we have to calculate the moles of Sn²⁺ that react.
![\frac{0.1015molSn^{2+} }{1L} .13.28 \times 10^{-3} L=1.348\times 10^{-3}molSn^{2+}](https://tex.z-dn.net/?f=%5Cfrac%7B0.1015molSn%5E%7B2%2B%7D%20%7D%7B1L%7D%20.13.28%20%5Ctimes%2010%5E%7B-3%7D%20L%3D1.348%5Ctimes%2010%5E%7B-3%7DmolSn%5E%7B2%2B%7D)
We also know the following relations:
- According to the balanced equation, 1 mole of Sn²⁺ reacts with 2 moles of Fe³⁺.
- 1 mole of Fe³⁺ is oxidized from 1 mole of Fe.
- The molar mass of Fe is 55.84 g/mol.
Then, for 1.348 × 10⁻3 moles of Sn²⁺:
![1.348\times 10^{-3}molSn^{2+}.\frac{2molFe^{3+} }{1molSn^{2+} } .\frac{1molFe}{1molFe^{3+} } .\frac{55.84gFe}{1molFe} =0.1505gFe](https://tex.z-dn.net/?f=1.348%5Ctimes%2010%5E%7B-3%7DmolSn%5E%7B2%2B%7D.%5Cfrac%7B2molFe%5E%7B3%2B%7D%20%7D%7B1molSn%5E%7B2%2B%7D%20%7D%20.%5Cfrac%7B1molFe%7D%7B1molFe%5E%7B3%2B%7D%20%7D%20.%5Cfrac%7B55.84gFe%7D%7B1molFe%7D%20%3D0.1505gFe)
If there are 0.1505 g of Fe in a 0.1875 g sample, the mass percentage of Fe is:
![\frac{0.1505g}{0.1875g} \times 100 \% = 80.27\%](https://tex.z-dn.net/?f=%5Cfrac%7B0.1505g%7D%7B0.1875g%7D%20%5Ctimes%20100%20%5C%25%20%3D%2080.27%5C%25)
The quick deployment of the bag which I think is quite similar to a cushion and a balloon. It holds your head/face away from any hard surfaces. Although it does not protect your legs or other extremities that are out of the bags range.
Answer:
130 Liters
Explanation:
if 1 mol is 22.4 L, then 5.8 mol is 130 L (129.92 but use sig figs)
Answer:
![m=0.0158g](https://tex.z-dn.net/?f=m%3D0.0158g)
Explanation:
Hello there!
In this case, it is possible to comprehend these mass-particles problems by means of the concept of mole, molar mass and the Avogadro's number because one mole of any substance has 6.022x10²³ particles and have a mass equal to the molar mass.
In such a way, for C₆H₁₂O₆, whose molar mass is about 180.16 g/mol, the referred mass would be:
![m=5.28x10^{19}molecules*\frac{1mol}{6.022x10^{23}molecules}*\frac{180.16g}{1mol}\\\\m=0.0158g](https://tex.z-dn.net/?f=m%3D5.28x10%5E%7B19%7Dmolecules%2A%5Cfrac%7B1mol%7D%7B6.022x10%5E%7B23%7Dmolecules%7D%2A%5Cfrac%7B180.16g%7D%7B1mol%7D%5C%5C%5C%5Cm%3D0.0158g)
Best regards!