Answer:
C₄H₂N₂
Explanation:
First we<u> calculate the moles of the gas</u>, using PV=nRT:
P = 2670 torr ⇒ 2670/760 = 3.51 atm
V = 300 mL ⇒ 300/1000 = 0.3 L
T = 228 °C ⇒ 228 + 273.16 = 501.16 K
- 3.51 atm * 0.3 L = n * 0.082atm·L·mol⁻¹·K⁻¹ * 501.16 K
Now we<u> calculate the molar mass of the compound</u>:
- 2.00 g / 0.0256 mol = 78 g/mol
Finally we use the percentages given to<em> </em><u>calculate the empirical formula</u>:
- C ⇒ 78 g/mol * 61.5/100 ÷ 12g/mol = 4
- H ⇒ 78 g/mol * 2.56/100 ÷ 1g/mol = 2
- N ⇒ 78 g/mol * 35.9/100 ÷ 14g/mol = 2
So the empirical formula is C₄H₂N₂
<span>ATP,O2 and NADPH are the </span>products<span>. H2O,NADP,ADP and Pi are the reactants. acts as an electron carrier between the cytochrome b6f and </span>photosystem 1 (PS1) complexes in the photosynthetic electron-transfer chain.
Photosystem II<span> (or water-plastoquinone oxidoreductase) is the first protein complex in the light-dependent reactions of oxygenic photosynthesis. It is located in the thylakoid membrane of plants, algae, and cyanobacteria.</span>
Answer:
![[Ag^{+}]=4.2\times 10^{-2}M](https://tex.z-dn.net/?f=%5BAg%5E%7B%2B%7D%5D%3D4.2%5Ctimes%2010%5E%7B-2%7DM)
Explanation:
Given:
[AgNO3] = 0.20 M
Ba(NO3)2 = 0.20 M
[K2CrO4] = 0.10 M
Ksp of Ag2CrO4 = 1.1 x 10^-12
Ksp of BaCrO4 = 1.1 x 10^-10

![Ksp=[Ba^{2+}][CrO_{4}^{2-}]](https://tex.z-dn.net/?f=Ksp%3D%5BBa%5E%7B2%2B%7D%5D%5BCrO_%7B4%7D%5E%7B2-%7D%5D)
![1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]](https://tex.z-dn.net/?f=1.2%5Ctimes%2010%5E%7B-10%7D%3D%280.20%29%5BCrO_%7B4%7D%5E%7B2-%7D%5D)
![[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}](https://tex.z-dn.net/?f=%5BCrO_%7B4%7D%5E%7B2-%7D%5D%3D%5Cfrac%7B1.2%5Ctimes%2010%5E%7B-10%7D%7D%7B%280.20%29%7D%3D%206.0%5Ctimes%2010%5E%7B-10%7D)
Now,

![Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]](https://tex.z-dn.net/?f=Ksp%3D%5BAg%5E%7B%2B%7D%5D%5E%7B2%7D%5BCrO_%7B4%7D%5E%7B2-%7D%5D)
![1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})](https://tex.z-dn.net/?f=1.1%5Ctimes%2010%5E%7B-12%7D%3D%5BAg%5E%7B%2B%7D%5D%5E%7B2%7D%5D%286.0%5Ctimes%2010%5E%7B-10%7D%29)
![[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}](https://tex.z-dn.net/?f=%5BAg%5E%7B%2B%7D%5D%5E%7B2%7D%5D%3D%5Cfrac%7B1.1%5Ctimes%2010%5E%7B-12%7D%7D%7B%286.0%5Ctimes%2010%5E%7B-10%7D%29%7D%3D%201.8%5Ctimes%2010%5E%7B-3%7D)
![[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M](https://tex.z-dn.net/?f=%5BAg%5E%7B%2B%7D%5D%3D%5Csqrt%7B1.8%5Ctimes%2010%5E%7B-3%7D%7D%3D4.2%5Ctimes%2010%5E%7B-2%7DM)
So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M
Answer:
D) Adding a catalyst
Explanation:
Adding a catalyst decreases activation energy and allows the reaction to occur more easily.
Answer:
1.3223 acres
Explanation:
a football field's area is 360 feet (120 yards) x 160 feet (53.333 yards) = 57,600 sq. feet
if the area of an acre is 43,560 sq. feet, then a football field in acres = 57,600 sq. feet / 43,560 sq. feet = 1.3223 acres
we can verify our answer by doing the same calculation in sq. yards:
football field = 120 yards x 53.33 yards = 6,400 sq. yards
an acre is 4,840 sq. yards
football field in acres = 6,400 sq. yards / 4,840 sq. yards = 1.3223 acres