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Leya [2.2K]
2 years ago
9

A student measures 10 g of baking soda and mixes it with 5 grams of vinegar. The mixture begins to bubble and fizz and feels col

d to the touch. Once the reaction is over the student weighs the solution and discovers that the resulting mixture weighs only 12g. What happened to the missing 3 grams of product? Use the Law of Conservation of Matter in your answer.
Chemistry
1 answer:
Tomtit [17]2 years ago
5 0

Answer:

Some of the product is a gas that goes to the atmosphere.

Explanation:

Based on the chemical reaction of baking soda (Sodium bicarbonate: NaHCO₃) with vinegar (Acetic acid: CH₃COOH):

NaHCO₃(aq) + CH₃COOH(aq) → H₂O(aq) +  CO₂(g) + CH₃COONa(aq)

When the reaction occurs, one of the products (CO₂) is a gas that will disappear of the mixture to stay in the atmosphere. That is the reason why there are missing 3 grams.

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You need to make 10.0 L of 1.2 M KNO3. What molar ( concentration) would the potassium nitrate solution need to be if you were t
solniwko [45]

Answer:

2.5L [NaCl] concentrate needs to be 4.8 Molar solution before dilution to prep 10L of 1.2M KNO₃ solution.

Explanation:

Generally, moles of solute in solution before dilution must equal moles of solute after dilution.

By definition Molarity = moles solute/volume of solution in Liters

=> moles solute = Molarity x Volume (L)

Apply moles before dilution = moles after dilution ...

=> (Molarity X Volume)before dilution = (Molarity X Volume)after dilution

=> (M)(2.5L)before = (1.2M)(10.0L)after

=> Molarity of 2.5L concentrate = (1.2M)(10.0L)/(2.5L) = 4.8 Molar concentrate

6 0
3 years ago
What volume of so2 is produced at 325 k and 1.35 atm when 15.0 grams of hcl reacts with excess k2so3?
Helen [10]
The balanced equation for the above reaction is as follows;
2HCl + K₂SO₃ ---> 2KCl + H₂O + SO₂
stoichiometry of HCl to SO₂ is 2:1
number of moles of HCl reacted - 15.0 g / 36.5 g/mol = 0.411 mol 
according to molar ratio 
number of SO₂ moles formed - 0.411 mol /2 = 0.206 mol
since we know the number of moles we can find volume using ideal gas law equation 
PV = nRT
where
 P - pressure - 1.35 atm x 101 325 Pa/atm = 136 789 Pa 
V - volume 
n - number of moles - 0.206 mol 
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 325 K
substituting values in the equation 

136 789 Pa x V = 0.206 mol x 8.314 Jmol⁻¹K⁻¹ x 325 K 
V = 4.07 L 
volume of SO₂ formed is 4.07 L


8 0
3 years ago
The C=C bond in 2-cyclohexenone produces an unusually strong signal. Explain using resonance structuresBased on the above resona
frosja888 [35]

Answer:

When two single single bonds separated by a double bond (e.g C=C-C=C or C=C-C=O in the case of 2-cyclohexenone), the effect of resonance among those there bonds will be observed.

Explanation:

Since the Oxygen atom has higher electronegativity, it will cause the electrons in the resonance bonds 'flow' toward the Oxygen atom, so that the C=C will 'lose' some electron. The signal read for that bond will be different from other alkene structure.

Attachment is the resonance structure of 2-cyclohexene.

8 0
3 years ago
If the distance between two objects is decreased to - of the original
Charra [1.4K]

Answer:

The new force will be \frac{1}{100} of the original force.

Explanation:

In the context of this problem, we're dealing with the law of gravitational attraction. The law states that the gravitational force between two object is directly proportional to the product of their masses and inversely proportional to the square of a distance between them.

That said, let's say that our equation for the initial force is:

F = G\frac{m_1m_2}{R^2}The problem states  that  the distance decrease to 1/10 of the original distance, this means:[tex]R_2 = \frac{1}{10}R

And the force at this distance would be written in terms of the same equation:

F_2 = G\frac{m_1m_2}{R_2^2}

Find the ratio between the final and the initial force:

\frac{F_2}{F} = \frac{G\frac{m_1m_2}{R_2^2}}{G\frac{m_1m_2}{R^2}}

Substitute the value for the final distance in terms of the initial distance:

\frac{F_2}{F} = \frac{G\frac{m_1m_2}{(\frac{R}{10})^2}}{G\frac{m_1m_2}{R^2}}

Simplify:

\frac{F_2}{F} = \frac{\frac{1}{100R^2}}{\frac{1}{R^2}}=\frac{1}{100}

This means the new force will be \frac{1}{100} of the original force.

8 0
3 years ago
If you were to take several random samples from a pure substance the samples
Nataliya [291]
Would all have the same ratio of elements.
3 0
3 years ago
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