Answer: 60.1K
Explanation:
Initial volume of gas V1 = 423.3mL
Initial temperature T1 = 49.2°C
Convert temperature in Celsius to Kelvin
( 49.2°C + 273 = 322.2K)
Final temperature T2 = ?
Final volume V2 = 79mL
According to Charle's law, the volume of a fixed mass of a gas is directly proportional to the temperature.
Mathematically, Charles' Law is expressed as: V1/T1 = V2/T2
423.3mL/322.2 = 79mL/T2
To get the value of T2, cross multiply
423.3mL x T2 = 322.2K x 79mL
423.3mL x T2 = 25453.8
T2 = (25453.8/423.3mL)
T2 = 60.1K
Thus, the new temperature of the gas is 60.1K
Answer:
Percent error = 1.5%
Explanation:
Given data:
Measured value of density of graphite = 2.3 g/cm³
Percent error = ?
Solution:
Formula:
Percent error = [Measured value - Actual value / actual value] × 100
Actual/accepted value of density of graphite = 2.266 g/cm³
Now we will put the values:
Percent error = [2.3 g/cm³ - 2.266 g/cm³ / 2.266 g/cm³] × 100
Percent error = [0.034 g/cm³ / 2.266 g/cm³] × 100
Percent error = 0.015 × 100
Percent error = 1.5%
Answer:
at the beginning:
pH = 0.745
Explanation:
HCl is a strong acid, so:
0.18 M 0.18 0.18.....equilibrium
before base is added:
∴ [ H3O+ ] ≅ <em>C </em>HCl = 0.18 M
⇒ pH = - Log [ H3O+ ] = - Log ( 0.18 )
⇒ pH = 0.745
The balanced chemical reaction:
<span>Cu + 2AgNO3 = Cu(NO3)2 + 2Ag
</span>
We are given the amount of the reactants to be used for the reaction. These values will be the starting point of our calculations.
9.85 g Cu ( 1 mol Cu / 63.55 g Cu ) = 0.15 mol Cu
31.0 g AgNO3 ( 1 mol AgNO3 / 169.87 g AgNO3 ) = 0.18 mol AgNO3
The limiting reactant is AgNO3.
0.18 mol AgNO3 ( 1 mol Cu(NO3)2 / 2 mol AgNO3 ) (187.56 g / 1 mol) =16.88 g Cu(NO3)2
0.15 mol Cu - 0.18 mol AgNO3 ( 1 mol Cu / 2 mol AgNo3) = 0.06 mol Cu excess
<span>0.06 mol Cu ( 63.55 g Cu / 1 mol Cu ) = 3.81 g Cu excess</span>
The answer is: the mass of carbon is 420.6 grams.
m(C₈H₁₈) = 500 g; mass of octane.
M(C₈H₁₈) = 114.22 g/mol; molar mass of octane.
n(C₈H₁₈) = m(C₈H₁₈) ÷ M(C₈H₁₈).
n(C₈H₁₈) = 500 g ÷ 114.22 g/mol.
n(C₈H₁₈) = 4.38 mol; amount of octane.
In one molecule of octane, there are eight carbon atoms:
n(C) = 8 · n(C₈H₁₈).
n(C) = 8 · 4.38 mol.
n(C) = 35.02 mol; amount of carbon.
m(C) = 35.02 mol · 12.01 g/mol.
m(C) = 420.6 g; mass of carbone.
