The pond weeds are the producers because they are a plant
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
Here, we are required to determine the volume of the earth which is 1.08326 × 10¹² km³ in liters.
<em>The volume of the earth is approximately</em>,
, 1.08326 × 10²⁴ liters
By conversion factors;
- <em>1dm³ = 1liter</em>
- However; <em>1km = 10000dm = 10⁴ </em><em>dm</em>
- Therefore, 1km³ = (10⁴)³ dm³.
Consequently, 1km³ = 10¹²dm³ = 10¹²liters.
The conversion factor from 1km³ to liters is therefore, c.f = 10¹²liters/km³
Therefore, the volume of the earth which is approximately, 1.08326 × 10¹² km³ can be expressed in liters as;
<em>1.08326 × 10¹² km³ × 10¹²liters/km³ </em>
The volume of the earth is approximately,
1.08326 × 10²⁴ liters.
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Answer:
0.382 atm
Explanation:
In order to find the pressure, you need to know the moles of carbon dioxide (CO₂) gas. This can be found by multiplying the mass (g) by the molar mass (g/mol) of CO₂. It is important to arrange the conversion in a way that allows for the cancellation of units.
Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)
Molar Mass (CO₂): 44.007 g/mol
15 grams CO₂ 1 mole
---------------------- x ------------------------ = 0.341 moles CO₂
44.007 grams
To find the pressure, you need to use the Ideal Gas Law equation.
PV = nRT
In this equation,
-----> P = pressure (atm)
-----> V = volume (L)
-----> n = moles
-----> R = Ideal Gas Constant (0.08206 atm*L/mol*K)
-----> T = temperature (K)
After you convert Celsius to Kelvin, you can plug the given and calculated values into the equation and simplify to find the pressure.
P = ? atm R = 0.08206 atm*L/mol*K
V = 20 L T = 0 °C + 273.15 = 273.15 K
n = 0.341 moles
PV = nRT
P(20 L) = (0.341 moles)(0.08206 atm*L/mol*K)(273.15 K)
P(20 L) = 7.64016
P = 0.382 atm