<u>Answer:</u> The mass of sulfur dioxide gas at STP for given amount is 16.8 g
<u>Explanation:</u>
At STP conditions:
22.4 L of volume is occupied by 1 mole of a gas.
So, 5.9 L of volume will be occupied by = ![\frac{1mol}{22.4L}\times 5.9L=0.263mol](https://tex.z-dn.net/?f=%5Cfrac%7B1mol%7D%7B22.4L%7D%5Ctimes%205.9L%3D0.263mol)
Now, to calculate the mass of a substance, we use the equation:
![\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B%5Ctext%7BMolar%20mass%7D%7D)
Moles of sulfur dioxide gas = 0.263 mol
Molar mass of sulfur dioxide gas = 64 g/mol
Putting values in above equation, we get:
![0.263mol=\frac{\text{Mass of sulfur dioxide gas}}{64g/mol}\\\\\text{Mass of sulfur dioxide gas}=(0.263mol\times 64g/mol)=16.8g](https://tex.z-dn.net/?f=0.263mol%3D%5Cfrac%7B%5Ctext%7BMass%20of%20sulfur%20dioxide%20gas%7D%7D%7B64g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20sulfur%20dioxide%20gas%7D%3D%280.263mol%5Ctimes%2064g%2Fmol%29%3D16.8g)
Hence, the mass of sulfur dioxide gas at STP for given amount is 16.8 g
Answer:
Option D. 53 moles.
Explanation:
The following data were obtained from the question:
Number of mole of C5H10O2 = 5.3 moles
Number of mole of Hydrogen in 5.3 moles of C5H10O2 =?
From the chemical formula of propyl acetate, C5H10O2,
1 mole of C5H10O2 contains 10 moles of H.
Therefore, 5.3 moles of C5H10O2 will contain = 5.3 × 10 = 53 moles of H.
Thus, 5.3 moles of C5H10O2 contains 53 moles of H.
Answer:
Au
Explanation:
For the density of a face-centered cubic:
![Density = \dfrac{4 \times M_w}{N_A \times a^3}](https://tex.z-dn.net/?f=Density%20%3D%20%5Cdfrac%7B4%20%5Ctimes%20M_w%7D%7BN_A%20%5Ctimes%20a%5E3%7D)
where
= molar mass of the compound
avogadro's constant
the volume of a unit cell
Given that:
Density
= 19.30 g/cm³
a = 0.408 nm
a = ![0.408 \times 10^{-9} \times 10^{2} \ cm](https://tex.z-dn.net/?f=0.408%20%5Ctimes%2010%5E%7B-9%7D%20%5Ctimes%2010%5E%7B2%7D%20%5C%20cm)
a = ![4.08 \times 10^ {-8} \ cm](https://tex.z-dn.net/?f=4.08%20%5Ctimes%2010%5E%20%7B-8%7D%20%5C%20cm)
∴
![19.3 = \dfrac{4 \times M_w}{(6.023 \tmes 10^{23})\times (4.08 \times 10^{-8})^3}](https://tex.z-dn.net/?f=19.3%20%3D%20%5Cdfrac%7B4%20%5Ctimes%20M_w%7D%7B%286.023%20%5Ctmes%2010%5E%7B23%7D%29%5Ctimes%20%284.08%20%5Ctimes%2010%5E%7B-8%7D%29%5E3%7D)
![M_w= \dfrac{19.3\times (6.023 \times 10^{23})\times (4.08 \times 10^{-8})^3}{4}](https://tex.z-dn.net/?f=M_w%3D%20%5Cdfrac%7B19.3%5Ctimes%20%286.023%20%5Ctimes%2010%5E%7B23%7D%29%5Ctimes%20%284.08%20%5Ctimes%2010%5E%7B-8%7D%29%5E3%7D%7B4%7D)
![M_w=197.37 \ g/mol](https://tex.z-dn.net/?f=M_w%3D197.37%20%5C%20g%2Fmol)
Thus, the molar mass of 197.37 g/mol element is Gold (Au).
The answer is (4) at the cathode, where reduction occurs. The Na+ gains one electron and become Na(l). So the reaction occurs at cathode and is reduction reaction.