1 mole of N2 produces 2 moles of NH3
OR...
14 x 2 grams of N2 produces 2(14 +3) grams of NH3
1 gram of N2 produces 34/28 grams of NH3
therefore, 56 grams produce (34/28 )x 56 =68 grams of NH3
the answer thus would be 68 grams of NH3
Answer:
The ionization equation is
⇄
(1)
Explanation:
The ionization equation is
⇄
(1)
As the Bronsted definition sais, an acid is a substance with the ability to give protons thus, H2PO4 is the acid and HPO42- is the conjugate base.
The Ka expression is the ratio between the concentration of products and reactants of the equilibrium reaction so,
![Ka = \frac{[HPO_{4}^{-2}] [H_{3}O^{+}]}{[H_{2}PO_{4}^{-}] [H_{2}O]} = 6.2x10^{-8}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BHPO_%7B4%7D%5E%7B-2%7D%5D%20%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5BH_%7B2%7DPO_%7B4%7D%5E%7B-%7D%5D%20%5BH_%7B2%7DO%5D%7D%20%3D%206.2x10%5E%7B-8%7D)
The pKa is

The pKa of H2CO3 is 6,35, thus this a stronger acid than H2PO4. The higher the pKa of an acid greater the capacity to donate protons.
In the body H2CO3 is a more optimal buffer for regulating pH due to the combination of the two acid-base equilibriums and the two pKa.
If the urine is acidified, according to Le Chatlier's Principle the equilibrium (1) moves to the left neutralizing the excess proton concentration.
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.
The volume of the gas at STP = 35.01 L
<h3>Further explanation</h3>
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure).
In general, the gas equation can be written

where
P = pressure, atm
V = volume, liter
n = number of moles
R = gas constant = 0.08206 L.atm / mol K
T = temperature, Kelvin
V=17.4 L
T = 23 + 273 = 296 K
P = 2.18 atm

The volume of the gas occupy at STP :

Answer:
The final temperature of the water will be 328.81 K .
Explanation:
Using the equation, <em>q = mcΔT</em>
here, <em>q = energy</em>
<em>m= mass</em>
<em>c= specific heat capacity </em>
<em>ΔT= change in temperature</em>
<em>Mass of water = 1kg (1000 g ) per liter</em>
<em>∴ 6.2 Liter of water = 6200 g</em>
<em>c of water ≈ 4.18 J /g/K</em>
<em>Now, </em>
<em>980000 = 6200*4.18*ΔT</em>
<em>ΔT = 37.81 K</em>
<em>∴ final temperature of the water = 291 + 37.81 = 328.81 K</em>