Define the term Distance.
2 answers:
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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?
Answer:
the block reaches higher than the sphere
\frac{y_{sphere}} {y_block} = 5/7
Explanation:
We are going to solve this interesting problem
A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp
Let's use the concept of conservation of energy
starting point. At the top of the ramp
Em₀ = U = m g y₁
final point. At the exit of the ramp
Em_f = K + U = ½ m v² + ½ I w² + m g y₂
notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂
energy is conserved
Em₀ = Em_f
m g y₁ = ½ m v² + ½ I w² + m g y₂
angular and linear velocity are related
v = w r
the moment of inertia of a sphere is
I = m r²
we substitute
m g (y₁ - y₂) = ½ m v² + ½ ( m r²) (
)²
m g h = ½ m v² (1 + )
where h is the difference in height between the two sides of the ramp
h = y₂ -y₁
mg h = (
m v²)
v = √5/7 √2gh
This is the exit velocity of the vertical movement of the sphere
v_sphere = 0.845 √2gh
B) is the same case, but for a box without friction
starting point
Em₀ = U = mg y₁
final point
Em_f = K + U = ½ m v² + m g y₂
Em₀ = Em_f
mg y₁ = ½ m v² + m g y₂
m g (y₁ -y₂) = ½ m v²
v = √2gh
this is the speed of the box
v_box = √2gh
to know which body reaches higher in the air we can use the kinematic relations
v² = v₀² - 2 g y
at the highest point v = 0
y = vo₀²/ 2g
for the sphere
y_sphere = 5/7 2gh / 2g
y_esfera = 5/7 h
for the block
y_block = 2gh / 2g
y_block = h
therefore the block reaches higher than the sphere