Answer:
0.486 L
Explanation:
Step 1: Write the balanced reaction
2 KCIO₃(s) ⇒ 2 KCI (s) + 3 O₂(g)
Step 2: Calculate the moles corresponding to 1.52 g of KCIO₃
The molar mass of KCIO₃ is 122.55 g/mol.
1.52 g × 1 mol/122.55 g = 0.0124 mol
Step 3: Calculate the moles of O₂ produced from 0.0124 moles of KCIO₃
The molar ratio of KCIO₃ to O₂ is 2:3. The moles of O₂ produced are 3/2 × 0.0124 mol = 0.0186 mol
Step 4: Calculate the volume corresponding to 0.0186 moles of O₂
0.0186 moles of O₂ are at 37 °C (310 K) and 0.974 atm. We can calculate the volume of oxygen using the ideal gas equation.
P × V = n × R × T
V = n × R × T/P
V = 0.0186 mol × (0.0821 atm.L/mol.K) × 310 K/0.974 atm = 0.486 L
Answer:
4.5g/mL
Explanation:
Given parameters:
Mass of ball = 36g
Volume of the ball = 8mL
Unknown:
Density of the ball = ?
Solution:
Density is the mass per unit volume of a substance.
Density = 
So;
Density = 
= 4.5g/mL
In your hand, the ball has higher potential energy than kinetic because it is still off of the ground but it isn't moving so there is no kinetic. As the ball rises, its potential and kinetic energy increases. At its peak, it has very high potential energy and very low kinetic energy. As it falls, the potential energy decreases but kinetic does not.
answer: A not sure but i hope it helps :) success
Answer:
The limiting reactant is the covers.
We are left with 3750 sheets of lined paper, 1125 sheets of graph paper and 25 staples.
Explanation:
the reaction is:
B - book
C - cover
Sl = sheet of lined paper
Sg - sheet of graph paper
St - staple
1B = 2C + 50Sl + 25Sg + 3St
We have covers for 150/2 = 75 books
We have lined paper for 7500/50 = 150 books
We have graph paper for 3000/25 = 120books
We have Staples for 250/3 = 83.(3) books, so for 83 books as we cannot make 1/3 of a book.
Because we only have covers for 75 books, covers are the limiting reactant.
After making 75 books we are left with:
0 covers
7500 - 75*50 = 3750 sheets of lined paper
3000 - 75*25 = 1125 sheets of graph paper
250 - 75*3 = 25 staples
