Answer:
0.4444 g/cm³ ≅ 0.44 g/cm³ (2 significant figures).
Explanation:
<em>d = m/V,</em>
where, d is the density of the material (g/cm³).
m is the mass of the material (m = 28 g).
V is the volume of the material (V = 63.0 cm³).
<em>∴ d = m/V </em>= (28 g)/(63.0 cm³) = <em>0.4444 g/cm³ ≅ 0.44 g/cm³ (2 significant figures).</em>
Answer is: the absolute pressure of the air in the balloon is 1.015 atm (102.84 kPa).
n = 0.250 mol; amount of substance.
V = 6.23 L; volume of the balloon.
T = 35°C = 308.15 K; temperature.
R = 0.08206 L·atm/mol·K, universal gas constant.
Ideal gas law: p·V = n·R·T.
p = n·R·T / V.
p = 0.250 mol · 0.08206 L·atm/mol·K · 308.15 K / 6.23 L.
p = 1.015 atm; presure of the air.
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.
Answer:
You may need better soil and a more plentiful amount of water coming from another source, and maybe find another way to contain the rain water
Explanation:you may need to draw it on paper sorry bout how it looks
The empirical formula is P₂O₃
