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:
[Ag⁺] = 5.0x10⁻¹⁴M
Explanation:
The product solubility constant, Ksp, of the insoluble salts PbI₂ and AgI is defined as follows:
Ksp(PbI₂) = [Pb²⁺] [I⁻]² = 1.4x10⁻⁸
Ksp(AgI) = [Ag⁺] [I⁻] = 8.3x10⁻¹⁷
The PbI₂ <em>just begin to precipitate when the product [Pb²⁺] [I⁻]² = 1.4x10⁻⁸</em>
<em />
As the initial [Pb²⁺] = 0.0050M:
[Pb²⁺] [I⁻]² = 1.4x10⁻⁸
[0.0050] [I⁻]² = 1.4x10⁻⁸
[I⁻]² = 1.4x10⁻⁸ / 0.0050
[I⁻]² = 2.8x10⁻⁶
<h3>[I⁻] = 1.67x10⁻³</h3><h3 />
So, as the [I⁻] concentration is also in the expression of Ksp of AgI and you know concentration in solution of I⁻ = 1.67x10⁻³M:
[Ag⁺] [I⁻] = 8.3x10⁻¹⁷
[Ag⁺] [1.67x10⁻³] = 8.3x10⁻¹⁷
<h3>[Ag⁺] = 5.0x10⁻¹⁴M</h3>
Precipitation occurs when the product of the ion concentration exceeds the Ksp.
Moles of water atoms = mass/molecular weight = 105/18 = 5.83 mol. Number of moles of hydrogen in water = 2 x moles of water = 11.66. Number of H atoms in water = moles of hydrogen x 6.02 x 10^23 = 7.019 x 10^24 ~ 7.02 x 10^24 atoms. Hope this helps.
Answer:
The coefficients are 6, 1, 3
Explanation:
HNCO →C3N3(NH2)3 + CO2
From the above equation, there are a total of 6 atoms of nitrogen on the right side and 1atom on the left. It can be balance by putting 6 in front of HNCO as shown below:
6HNCO → C3N3(NH2)3 + CO2
Now there are 6 atoms of carbon on the left side and 4 atoms on the right side. It can be balance by putting 3 in front of CO2 as shown below:
6HNCO → C3N3(NH2)3 + 3CO2
Now the equation is balanced as the numbers of atoms of the different elements on both sides of the equation are the same.
The coefficients are 6, 1, 3
