C.) protons , the nucleus contains 2 subatomic particles , one protons & neutrons ( protons have a + charge & neutrons have no electrical charge
Answer:
The work done by the system is 100 J
Explanation:
Given details
The cross sectional area of the of the container is A = 100.0 cm^2 = 0.01m²
The total distance pushed by the piston is d = 10 cm = 0.10m
The total external pressure by which piston pushed is P = 100 kPa
From above data, the following relation can be used to determine the change in volume of the container
∆V = A * d
∆V = 0.01 * 0.10 = 0.001 m³
By using the following relation, the work done by the system is calculated as;
Work done W = P * ∆V
W = 100 * 0.001 = 0.1 kJ = 100 J
The work done by the system is 100 J
The 6 M solution NaOH solution diluted before titrating the vinegar sample because pH at the equivalence point changes very rapidly . It is very useful to have dilution solution , so that number of moles that are transferred per drop of the solution is low . So, pH change smaller amount per drop. Having more dilute solution also allows the titration to use volume that is easier to work with.
If you used the 6 M solution was used then the volume required for titration would go down by a factor of 6/0.3 = 20
This means that if a titration took 20ml of the 0.3 M solution then it would take 1ml of the NaOH solution.
Titration consists of adding a controlled and known amount of standard solution with an unknown until the reaction to be complete.
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Answer: 0.11 g/ml
Explanation:
Half-life = 81 minutes
First we have to calculate the rate constant, we use the formula :
Now we have to calculate the age of the sample:
Expression for rate law for first order kinetics is given by:
where,
k = rate constant = 0.008\text{minutes}^{-1}[/tex]
t = time of decomposition = 324 minutes
a = let initial concentration of the reactant = 1.8 g/ml
a - x = concentration after decay process = ?
Now put all the given values in above equation, we get
Thus concentration after 324 minutes will be 0.11 g/ml.
Answer:
In this reaction HCl is the limiting reactant and aluminium is the reactant in excess.
Explanation:
Step 1: Data given
Mass of HCl = 69.0 grams
Mass of Al = 78.0 grams
Molar mass HCl = 36.46 g/mol
Atomic mass Al = 26.99 g/mol
Step 2: The balanced equation
6HCl(aq) + 2Al(s) → 3H2(g) + 2AlCl3(s)
Step 3: calculate moles HCl
Moles HCl = mass HCl / molar mass HCl
Moles HCl = 69.0 grams / 36.46 g/mol
Moles HCl = 1.89 moles
Step 4: Calculate moles Al
Moles Al = 78.0 grams / 26.99 g/mol
Moles Al = 2.89 moles
Step 5: Calculate the limiting reactant
For 6 moles HCl we need 2 moles Al to produce 3 moles H2 and 2 moles AlCl3
HCl is the limiting reactant. It will completely be consumed (1.89 moles). Aluminium is the reactant in excess. There will react 1.89 / 3 = 0.63 moles
There will remain 2.89 - 0.63 = 2.26 moles aluminium
In this reaction HCl is the limiting reactant and aluminium is the reactant in excess.
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