Answer:
Percent error = -4.7%
Explanation:
Given data:
Actual density value = 19.3 g/cm³
Measured value of density by student = 18.4 g/cm³
Percent error = ?
Solution:
Formula:
Percent error = ( measured value - actual value / actual value ) × 100
Now we will put the values.
Percent error = ( 18.4 g/cm³ - 19.3 g/cm³ / 19.3 g/cm³) × 100
Percent error = (-0.9 g/cm³ / 19.3 g/cm³) × 100
Percent error = -0.047 × 100
Percent error = -4.7%
Negative sign shows that measured value is less than actual value.
Answer:
0.03atm
Explanation:
Given parameters:
Total pressure = 780torr
Partial pressure of water vapor = 1.0atm
Unknown:
Partial pressure of radon = ?
Solution:
A sound knowledge of Dalton's law of partial pressure will help solve this problem.
The law states that "the total pressure of a mixture of gases is equal to the sum of the partial pressures of the constituent gases".
Mathematically;
P
= P
+ P
+ P
Since the total pressure is 780torr, convert this to atm;
760torr = 1 atm
780torr = 
atm = 1.03atm
For this problem;
Total pressure = Partial pressure of radon + Partial pressure of water vapor
1.03 = Partial pressure of radon + 1.0
Partial pressure of radon = 1.03 - 1.00 = 0.03atm
<h3>Answer:</h3>
Limiting reactant is Lithium
<h3>
Explanation:</h3>
<u>We are given;</u>
- Mass of Lithium as 1.50 g
- Mass of nitrogen is 1.50 g
We are required to determine the rate limiting reagent.
- First, we write the balanced equation for the reaction
6Li(s) + N₂(g) → 2Li₃N
From the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
- Second, we determine moles of Lithium and nitrogen given.
Moles = Mass ÷ Molar mass
Moles of Lithium
Molar mass of Li = 6.941 g/mol
Moles of Li = 1.50 g ÷ 6.941 g/mol
= 0.216 moles
Moles of nitrogen gas
Molar mass of Nitrogen gas is 28.0 g/mol
Moles of nitrogen gas = 1.50 g ÷ 28.0 g/mol
= 0.054 moles
- According to the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
- Therefore, 0.216 moles of lithium will require 0.036 moles (0.216 moles ÷6) of nitrogen gas.
- On the other hand, 0.054 moles of nitrogen, would require 0.324 moles of Lithium.
Thus, Lithium is the limiting reagent while nitrogen is in excess.
Answer:
[KHP] = 0.0428M
Explanation:
2 methods to calculate concentration after dilution
1. Use dilution equation
Molarity of concentrate (M₁) x Volume of Concentrate (V₁)
= Molarity of dilute (M₂) x Volume of dilute (V₂)
M₁ x V₁ = M₂ x V₂ => M₂ = M₁ x V₁ / V₂ = (1.07M)(10ml)/(250ml) = 0.0428M
2. Concentration Equation
moles KHPh = Molarity (M) x Volume (V) = 1.07M x 0.010L =0.0107 moles KHP
Concentration KHP = moles solute / volume of solution in Liters
= 0.0107 moles KHP / 0.25L = 0.0428M
