The partial pressure of hydrogen is 0.31 atm
calculation
find the number of hydrogen moles the container, that is
25/100 x 6.4 =1.6 moles of hydrogen
find the partial pressure for hydrogen in 1.6 moles
that is 6.4 moles= 1.24 atm
1.6 moles= ?
by cross multiplication
1.6moles x1.24 atm/ 6.4 moles= 0.31 atm
Answer:
8.3 kJ
Explanation:
In this problem we have to consider that both water and the calorimeter absorb the heat of combustion, so we will calculate them:
q for water:
q H₂O = m x c x ΔT where m: mass of water = 944 mL x 1 g/mL = 944 g
c: specific heat of water = 4.186 J/gºC
ΔT : change in temperature = 2.06 ºC
so solving for q :
q H₂O = 944 g x 4.186 J/gºC x 2.06 ºC = 8,140 J
For calorimeter
q calorimeter = C x ΔT where C: heat capacity of calorimeter = 69.6 ºC
ΔT : change in temperature = 2.06 ºC
q calorimeter = 69.60J x 2.06 ºC = 143.4 J
Total heat released = 8,140 J + 143.4 J = 8,2836 J
Converting into kilojoules by dividing by 1000 we will have answered the question:
8,2836 J x 1 kJ/J = 8.3 kJ
Answer:
filtration is the process of using a filter to remove solids from liquids or gasses.
Example:
an example of this is tea.
Answer:
6.61 Pounds
Solution:
Step 1: Calculate Mass of Water as;
Density = Mass ÷ Volume
Solving for Mass,
Mass = Density × Volume ------ (1)
As,
Density of Water = 1 g.cm⁻³
And,
3 L of Water = 3000 cm³
Putting values in equation 1,
Mass = 1 g.cm⁻³ × 3000 cm³
Mass = 3000 g
Step 2: Convert Grams into Pounds;
As,
1 Gram = 0.002204 Pounds
So,
3000 Grams = X Pounds
Solving for X,
X = (3000 Grams × 0.002204 Pounds) ÷ 1 Gram
X = 6.61 Pounds
Volume of Hydrogen V1 = 351mL
Temperature T1 = 20 = 20 + 273 = 293 K
Temperature T2 = 38 = 38 + 273 = 311 K
We have V1 x T2 = V2 x T1
So V2 = (V1 x T2) / T1 = (351 x 311) / 293 = 372.56
Volume at 38 C = 373 ml
