what is a fire piston?
*a tool for starting a fire, likely invented in Ancient Southeast Asia.*
dont know the second one lol.
Answer:
Boiling point
Explanation:
Distillation is one of the most widely used separation technique in chemistry. It is used to separate a mixture of liquid substances with different boiling point. Hence, the basis of the separation is BOILING POINT DIFFERENCE.
In the procedure, the liquid substances are heated until they turn gaseous, which they do at different times considering their different boiling points. The separated components are then converted back to liquid states in a process called CONDENSATION.
Answer:
oxygen is limiting reactant
Explanation:
Given data:
Mass of hydrogen = 16.7 g
Mass of oxygen = 15.4 g
Limiting reactant = ?
Solution:
Chemical equation:
2H₂ + O₂ → 2H₂O
Number of moles of hydrogen:
Number of moles = mass/ molar mass
Number of moles = 16.7 g/ 2 g/mol
Number of moles = 8.35 mol
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 15.4 g/ 32 g/mol
Number of moles = 0.48 mol
Now we will compare the moles of both reactant with product,
H₂ : H₂O
2 : 2
8.35 : 8.35
O₂ : H₂O
1 : 2
0.48 : 2×0.48 = 0.96 mol
The number of moles of water produced by oxygen are less so it will limiting reactant.
Answer:
10.5 g
Explanation:
Step 1: Given data
- Molar concentration of the solution (C): 0.243 M
- Volume of solution (V): 0.580 L
Step 2: Calculate the moles of solute (n)
Molarity is equal to the moles of solute divided by the liters of solution.
M = n/V
n = M × V
n = 0.243 mol/L × 0.580 L = 0.141 mol
Step 3: Calculate the mass corresponding to 0.141 moles of KCl
The molar mass of KCl is 74.55 g/mol.
0.141 mol × 74.55 g/mol = 10.5 g
Answer:
320 g
Step-by-step explanation:
The half-life of Co-63 (5.3 yr) is the time it takes for half of it to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows:
No. of Fraction Mass
half-lives t/yr Remaining Remaining/g
0 0 1
1 5.3 ½
2 10.6 ¼
3 15.9 ⅛ 40.0
4 21.2 ¹/₁₆
We see that 40.0 g remain after three half-lives.
This is one-eighth of the original mass.
The mass of the original sample was 8 × 40 g = 320 g
