Answer: It would be malleable, solids, luster, conductors, reactive
Explanation:
Answer: 5. Positive
6. Negative
9. Oxygen carbon hydrogen nitrogen
Explanation:
Answer:
Explanation:
<u>1) Data:</u>
a) V = 93.90 ml
b) T = 28°C
c) P₁ = 744 mmHg
d) P₂ = 28.25 mmHg
d) n = ?
<u>2) Conversion of units</u>
a) V = 93.90 ml × 1.000 liter / 1,000 ml = 0.09390 liter
b) T = 28°C = 28 + 273.15 K = 301.15 K
c) P₁ = 744 mmHg × 1 atm / 760 mmHg = 0.9789 atm
d) P₂ = 28.5 mmHg × 1 atm / 760 mmHg = 0.0375 atm
<u>3) Chemical principles and formulae</u>
a) The total pressure of a mixture of gases is equal to the sum of the partial pressures of each gas. Hence, the partical pressure of the hydrogen gas collected is equal to the total pressure less the vapor pressure of water.
b) Ideal gas equation: pV = nRT
<u>4) Solution:</u>
a) Partial pressure of hydrogen gas: 0.9789 atm - 0.0375 atm = 0.9414 atm
b) Moles of hygrogen gas:
pV = nRT ⇒ n = pV / (RT) =
n = (0.9414 atm × 0.09390 liter) / (0.0821 atm-liter /K-mol × 301.15K) =
n = 0.00358 mol (which is rounded to 3 significant figures) ← answer
Answer:
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Explanation:
The pH of a buffer solution is calculated using following relation
Thus the pH of buffer solution will be near to the pKa of the acid used in making the buffer solution.
The pKa value of HC₃H₅O₃ acid is more closer to required pH = 4 than CH₃NH₃⁺ acid.
pKa = -log [Ka]
For HC₃H₅O₃
pKa = 3.1
For CH₃NH₃⁺
pKa = 10.64
pKb = 14-10.64 = 3.36 [Thus the pKb of this acid is also near to required pH value)
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Half of the acid will get neutralized by the given base and thus will result in equal concentration of both the weak acid and the salt making the pH just equal to the pKa value.
Answer:
(a) The rate of formation of K2O is 0.12 M/s.
The rate of formation of N2 is also 0.12 M/s
(b) The rate of decomposition of KNO3 is 0.24 M/s
Explanation:
(a) From the equation of reaction, the mole ratio of K2O to O2 is 2:5.
Rate of formation of O2 is 0.3 M/s
Therefore, rate of formation of K2O = (2×0.3/5) = 0.12 M/s
Also from the equation of reaction, mole ratio of N2 to O2 is 2:5.
Rate of formation of N2 = (2×0.3/5) = 0.12 M/s
(b) From the equation of reaction, mole ratio of KNO3 to O2 is 4:5.
Therefore, rate of decomposition of KNO3 = (4×0.3/5) = 0.24 M/s
