Answer : The temperature of the gas is, 245.9 K
Explanation :
To calculate the temperature of gas we are using ideal gas equation:
where,
P = pressure of gas = 2770.96 torr = 3.646 atm
Conversion used : (1 atm = 760 torr)
V = volume of gas = 88.84 L
T = temperature of gas = ?
R = gas constant = 0.0821 L.atm/mole.K
w = mass of gas = 609.64 g
M = molar mass of gas = 38 g/mole
Now put all the given values in the ideal gas equation, we get:
Therefore, the temperature of the gas is, 245.9 K
Answer:
The correct answer is option C.
Explanation:
1.0 g sample of a cashew :
Heat released on combustion of 1.0 gram of cashew = -Q
We have mass of water = m = 1000 g
Specific heat of water = c = 4.184 J/g°C
ΔT = 30°C - 25°C = 5°C
Heat absorbed by the water : Q
Heat released on combustion of 1.0 gram of cashew is -20,920 J.
3.0 g sample of a marshmallows :
Heat released on combustion of 3.0 g sample of a marshmallows = -Q'
We have mass of water = m = 2000 g
Specific heat of water = c = 4.184 J/g°C
ΔT = 30°C - 25°C = 5°C
Heat absorbed by the water : Q'
Heat released on 3.0 g sample of a marshmallows= -Q' = -41,840 J
Heat released on 1.0 g sample of a marshmallows : q
Heat released on combustion of 1.0 gram of marshmallows -13,946.67 J.
-20,920 J. > -13,946.67 J
The combustion of 1.0 g of cashew releases more energy than the combustion of 1.0 g of marshmallow.
Answer:
* Before addition of any KOH:
pH = 0,0301
*After addition of 25.0 mL KOH:
pH = 1,30
*After addition of 50.0 mL KOH:
pH = 2,87
*After addition of 75.0 mL KOH:
pH = 6,70
*After addition of 100.0 mL KOH:
pH = 10,7
Explanation:
H₃PO₃ has the following equilibriums:
H₃PO₃ ⇄ H₂PO₃⁻ H⁺
k = [H₂PO₃⁻] [H⁺] / [H₃PO₃] k = 10^-(1,30) <em>(1)</em>
H₂PO₃⁻ ⇄ HPO₃²⁻ + H⁺
k = [HPO₃²⁻] [H⁺] / [H₂PO₃⁻] k = 10^-(6,70) <em>(2)</em>
Moles of H₃PO₃ are:
0,0500L×(1,8mol/L) = 0,09 moles of H₃PO₃
* Before addition of any KOH:
Using (1), moles in equilibrium are:
H₃PO₃: 0,09-x
H₂PO₃⁻: x
H⁺: x
Replacing:
4.51x10⁻³ - 0.050x -x² = 0
The right solution of x is:
x = 0.0466589
As volume is 0,050L
[H⁺] = 0.0466589moles / 0,050L = 0,933M
As pH = -log [H⁺]
<em>pH = 0,0301</em>
*After addition of 25.0 mL KOH:
0,025L×1,8M = 0,045 moles of KOH that reacts with H₃PO₃ thus:
KOH + H₃PO₃ → H₂PO₃⁻ + H₂O
That means moles of KOH will be the same of H₂PO₃⁻ and moles of H₃PO₃ are 0,09moles - 0,045moles = 0,045moles
Henderson-Hasselbalch formula is:
pH = pka + log₁₀ [A⁻] /[HA]
Where A⁻ is H₂PO₃⁻ and HA is H₃PO₃.
Replacing:
pH = 1,30 + log₁₀ [0,045mol] / [0,045mol]
<em>pH = 1,30</em>
*After addition of 50.0 mL KOH:
The addition of 50.0 mL KOH consume all H₃PO₃. Thus, in the solution you will have just H₂PO₃⁻. Thus, moles in solution for the equilibrium will be:
H₂PO₃⁻: 0,09-x
HPO₃²⁻: x
H⁺: x
Replacing:
1.8x10⁻⁸ - 2x10⁻⁷x - x² = 0
The right solution of x is:
x = 0.000134064
As volume is 50,0mL + 50,0mL = 100,0mL
[H⁺] = 0.000134064moles / 0,100L = 1.34x10⁻³M
As pH = -log [H⁺]
<em>pH = 2,87</em>
*After addition of 75.0 mL KOH:
Applying Henderson-Hasselbalch formula you will have 0,045 moles of both H₂PO₃⁻ HPO₃²⁻ and pka: 6,70:
pH = 6,70 + log₁₀ [0,045mol] / [0,045mol]
<em>pH = 6,70</em>
*After addition of 100.0 mL KOH:
You will have just 0,09moles of HPO₃²⁻, the equilibrium will be:
HPO₃²⁻ + H₂O ⇄ H₂PO₃⁻ + OH⁻ with kb = kw/ka = 1x10⁻¹⁴/10^-(6,70) = 5,01x10⁻⁸
kb = [H₂PO₃⁻] [OH⁻] / [HPO₃²⁻]
Moles are:
H₂PO₃⁻: x
OH⁻: x
HPO₃²⁻: 0,09-x
Replacing:
4.5x10⁻⁹ - 5.01x10⁻⁸x - x² = 0
The right solution of x is:
x = 0.000067057
As volume is 50,0mL + 100,0mL = 150,0mL
[OH⁻] = 0.000067057moles / 0,150L = 4.47x10⁻⁴M
As pH = 14-pOH; pOH = -log [OH⁻]
<em>pH = 10,7</em>
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I hope it helps!
Explanation:
Molarity of copper sulfate solution = 1.00 M
Volume of the copper sulfate solution = 50.0 mL = 0.050 L
Moles of copper sulfate = n
n = 0.050 L × 1.00 M= 0.050 mol
1 mol of copper sulfate has 1 mol of copper . Then 0.050 mol of copper sulfate has :
of copper
a) Mass of 0.050 moles of copper = 0.050 mol × 63.5 g/mol =3.175 g
b) The identity of the compound which formed after the reaction is copper hydroxide.
c) The complete equation for the reaction that occurs when copper sulfate and potassium hydroxide are mixed:
d) ..[1]
..[2]
Common ion both sides are removed. The net ionic equation is given as:
e) Volume of solution after mixing of both solution,V= 50 mL + 50ml = 100 mL
Mass of final solution ,m= 1 mL
Density of solution ,d= 1 g/mL (same as pure water)
Heat capacity of the solution = c = 4.186 J/g°C (same as pure water)
Change in temperature of the solution,ΔT = 27.7 °C- 21.5 °C=6.2°C
Enthalpy of the reaction = ΔH =
ΔH =
The ΔH for the reaction that occurs on mixing is 51.90 kJ/mol.