Answer:
![\mathbf{surface \ area} = \mathbf {64.5 cm^2}](https://tex.z-dn.net/?f=%5Cmathbf%7Bsurface%20%20%5C%20%20area%7D%20%3D%20%5Cmathbf%20%7B64.5%20cm%5E2%7D)
Explanation:
The missing diagram is attached in the file below:
The formula for calculating the surface of triangular prism is expressed as:
![\mathbf{surface \ area} = \mathbf {bh + (s_1 + s_2 + s_3)H}](https://tex.z-dn.net/?f=%5Cmathbf%7Bsurface%20%20%5C%20%20area%7D%20%3D%20%5Cmathbf%20%7Bbh%20%2B%20%28s_1%20%2B%20s_2%20%2B%20s_3%29H%7D)
where ;
base b = 28 mm = 2.8 cm
height h of the triangle = 3.75 cm
the sides of the triangle
= 2.8 , 4 and 4 respectively
the height if the prism H = 5 cm
![\mathbf{surface \ area} = \mathbf {(2.8*3.75)+ (0.28 + 4+ 4)5}](https://tex.z-dn.net/?f=%5Cmathbf%7Bsurface%20%20%5C%20%20area%7D%20%3D%20%5Cmathbf%20%7B%282.8%2A3.75%29%2B%20%280.28%20%2B%204%2B%204%295%7D)
![\mathbf{surface \ area} = \mathbf {(10.5)+ (8.28)5}](https://tex.z-dn.net/?f=%5Cmathbf%7Bsurface%20%20%5C%20%20area%7D%20%3D%20%5Cmathbf%20%7B%2810.5%29%2B%20%288.28%295%7D)
![\mathbf{surface \ area} = \mathbf {(10.5)+ (54)}](https://tex.z-dn.net/?f=%5Cmathbf%7Bsurface%20%20%5C%20%20area%7D%20%3D%20%5Cmathbf%20%7B%2810.5%29%2B%20%2854%29%7D)
![\mathbf{surface \ area} = \mathbf {64.5 cm^2}](https://tex.z-dn.net/?f=%5Cmathbf%7Bsurface%20%20%5C%20%20area%7D%20%3D%20%5Cmathbf%20%7B64.5%20cm%5E2%7D)
1 is the answer hope this helps you
Answer: 1424 grams
Explanation:
![2NaN_3\rightarrow 2a+3N_2](https://tex.z-dn.net/?f=2NaN_3%5Crightarrow%202a%2B3N_2)
![Density=\frac{mass}{Volume}](https://tex.z-dn.net/?f=Density%3D%5Cfrac%7Bmass%7D%7BVolume%7D)
![1.25g/L=\frac{mass}{736L}](https://tex.z-dn.net/?f=1.25g%2FL%3D%5Cfrac%7Bmass%7D%7B736L%7D)
![Mass=920g](https://tex.z-dn.net/?f=Mass%3D920g)
Thus mass of nitrogen produced is 920 g.
![\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)
![\text{Number of moles of nitrogen}=\frac{920g}{28g/mol}=32.8moles](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%20of%20nitrogen%7D%3D%5Cfrac%7B920g%7D%7B28g%2Fmol%7D%3D32.8moles)
According to stoichiometry:
3 mole of
are produced from= 2 mole of ![NaN_3](https://tex.z-dn.net/?f=NaN_3)
Thus 32.8 moles of
are produced from=
of ![NaN_3](https://tex.z-dn.net/?f=NaN_3)
Mass of ![NaN_3=moles\times {\text {molar mass}}=21.9\times 65=1424g](https://tex.z-dn.net/?f=NaN_3%3Dmoles%5Ctimes%20%7B%5Ctext%20%7Bmolar%20mass%7D%7D%3D21.9%5Ctimes%2065%3D1424g)
Thus 1424 grams of sodium azide is required to produce 736 L of Nitrogen gas with the density of 1.25 g/L.
Answer:
D
Explanation:
bothe cylinders are at room temperature so no cylinder has more themal energy. That crosses off the answers A B and C.