Answer:
29 counts per minute (29 cpm)
Explanation:
The half-life of a radioactive isotope is the time needed for the activity of the sample to halve.The half-life of Francium-223 is exactly 22 minutes: this means that its activity after 22 minutes becomes exactly half of its initial value. Since the initial activity of this sample of francium-223 was 58 cpm, the activity after 22 minutes will be
Height of Aeroplane = 500 meters
Speed of Aeroplane = 200 m/s
Acceleration due to gravity = g = 10m/s^2
time of package to reach the ground = t = ?
Horizontal distance = d = ?
<h2>_____________________________________</h2><h3>SOLUTION:</h3>When the bomb is dropped, it will have an initial horizontal velocity which is equal to the speed of the plane as the plane is just moving in horizontal direction. So the bomb fall and will travel forward.
Initial horizontal velocity is given by,
Initial vertical velocity is given by,
By the second equation of motion under gravity,
<u>For horizontal distance(d)</u>:
Horizontal distance has no acceleration thus d is given by
d = vt
d = 200x10
d = 2000 meters
<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>
The weight percentage of sodium carbonate in the mixture is = 67.71%
The weight percentage of sodium bicarbonate in the mixture is = 32.57%
<h3>What does "molar mass" ?</h3>The total mass throughout grams of all the atoms needed to form a molecule per mole is what makes up the molar mass, also known as the molecular weight. Grams per mole is the unit measuring molar mass.
<h3>According to the given information:</h3>The interaction between sodium carbonate and hydrochloric acid has the following equation:
Na₂CO₃ + HCl --> NaCl + CO₂ + H₂O
The reaction's balanced equation is,
Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O -------(2).
When sodium hydrogen carbonate as well as hydrochloric acid react, the following equation results.
NaCO₃ + HCl --> NaCl + CO₂ + H₂O ----------------(3)
The Molar mass :
Na₂CO₃ = 106g/mol
NaCO₃ = 84g/mol
NaCl = 58.5g/mol
The mass of the mixture is Na₂CO₃/NaCO₃ = 9.540g
The molar masses of the constituent elements that make up the compounds are added to determine the molar weight of the substances. In both chemical equations (2) and (3), the unitary technique is employed to calculate the mass of the mixture based on the number of moles (3).
The mass of the sodium carbonate and sodium bicarbonate is calculated with the help of the mass of NaCl formed by the mixture of Na₂CO₃/NaCO₃ with HCl.
The mixture has the following percentage of sodium carbonate:
% of Na₂CO₃ =
= (6.46/9.540)*100
= 67.71%
The weight percentage of sodium carbonate in the mixture is = 67.71%
The mixture has the following percentage of NaHCO3:
% of NaHCO₃ =
=
= (3.108/9.540)*100
= 32.57%
The weight percentage of sodium bicarbonate in the mixture is = 32.57%
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I understand that the question you are looking for is:
A mixture of sodium carbonate and sodium hydrogen carbonate is treated with aqueous hydrochloric acid. the unbalanced equations for the resulting's reactions are:
Na2CO3 (s) + HCl (aq) --> NaCl (aq) + CO2(g) + H2O (l)
NaHCO3(s) + HCl (aq) --> NaCl (aq) + CO2(g) + H2O (l)
You treat 9.540g of Na2CO3/NaHCO3 mixture with an excess of aqueous HCl and isolate 9.355g of NaCl. What is the weight percent of each substance in the mixture?