Answer:
Atomic number = 25 protons
Mass number = ( 25 + 30 ) = 55
The volume of your 0.25 m stock solution that is needed to make 200 ml of 0.010 m NaCl is <u>0.008 L</u>
Concentration is the abundance of a constituent divided by way of the overall volume of an aggregate. several sorts of mathematical descriptions may be outstanding: mass concentration, molar concentration, variety concentration, and extent awareness.
Calculation:-
C₁ = 0.25 M
V₁ = ?
C₂ = 0.010 M
V₂ = 200 ml = 0.2 L
V₁ = C₂V₂/C₁
= 0.010 × 0.2 / 0.25
=<u> 0.008 L</u>
The concentration of a substance is the quantity of solute found in a given amount of solution. Concentrations are normally expressed in terms of molarity, defined because of the variety of moles of solute in 1 L of answer.
The Concentration of an answer is a measure of the quantity of solute that has been dissolved in a given amount of solvent or answer. A concentrated answer is one that has a rather huge quantity of dissolved solute.
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The original material has decayed 75%
<h3>Further explanation</h3>
Given
two half-life
Required
The decayed sample
Solution
General formulas used in decay:

t = duration of decay
t 1/2 = half-life
N₀ = the number of initial radioactive atoms
Nt = the number of radioactive atoms left after decaying during T time
t = 2 x t 1/2
Input the value :
The amount of Nt = 25% No
Answer:
34.28 L ( 1.5*22.4 L)
Explanation:
Calculation of the moles of aluminum as:-
Mass = 55 g
Molar mass of aluminum = 26.981539 g/mol
The formula for the calculation of moles is shown below:
Thus,

According to the reaction:-

4 moles of aluminum react with 3 moles of oxygen gas
1 mole of aluminum react with
moles of oxygen gas
2.0384 moles of aluminum react with
moles of oxygen gas
Moles of oxygen gas = 1.5288 moles
At STP,
Pressure = 1 atm
Temperature = 273.15 K
Using ideal gas equation as:

where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
1 atm × V = 1.5288 mol × 0.0821 L.atm/K.mol × 273.15 K
⇒V = 34.28 L ( 1.5*22.4 L)
<span>The control is what you have to compare the rest of the experiment to.</span>