Physical property of matter because you don't have to change its chemical make-up to tell, you can observe it using your senses and tools.
Answer:
Explanation:
We are asked to find the volume of a solution given the moles of solute and molarity.
Molarity is a measure of concentration in moles per liter. It is calculated using the following formula:
We know there are 0.14 moles of potassium chloride (KCl), which is the solute. The molarity of the solution is 1.8 molar or 1.8 moles of potassium chloride per liter.
- moles of solute = 0.14 mol KCl
- molarity= 1.8 mol KCl/ L
- liters of solution=x
Substitute these values/variables into the formula.
We are solving for x, so we must isolate the variable. First, cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.
Now x is being multiplied by 1.8 moles of potassium chloride per liter. The inverse operation of multiplication is division, so we divide both sides by 1.8 mol KCl/L.
The units of moles of potassium chloride cancel.
The original measurements of moles and molarity have 2 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 7 in the ten-thousandth place tells us to round the 7 up to a 8.
There are approximately <u>0.078 liters of solution.</u>
Given:
Half life(t^ 1/2) :30 years
A0( initial mass of the substance): 200 mg.
Now we know that
A= A0/ [2 ^ (t/√t)]
Where A is the mass that remains after t years.
A0 is the initial mass
t is the time
t^1/2 is the half life
Substituting the given values in the above equation we get
A= [200/ 2^(t/30) ] mg
Thus the mass remaining after t years is [200/ 2^(t/30) ] mg
Answer:
The correct answer is 2.75 grams of HCl.
Explanation:
The given balanced equation is:
CaCO₃ (s) + 2HCl (aq) ⇒ CaCl₂ (aq) + H₂O (l) + CO₂ (g)
Based on the given information, one mole of calcium carbonate is reacting with two moles of HCl. The molecular mass of HCl is 36.5 grams, thus, the mass of 2 moles of HCl will be, 36.5 × 2 = 73 grams
The molecular mass of CaCO₃ is 100 gram per mole, that is, the mass of 1 mole of CaCO₃ is 100 grams, therefore, the mass of HCl required for reacting with 3.75 grams of CaCO₃ will be,
= 3.75 × 2 × 36.5 / 100 = 2.74 grams of HCl.
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