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alexgriva [62]
3 years ago
8

Julissa combined baking soda and calcium chloride solution in a beaker. she observed bubbling and a white solid formed. Julissa

completed the reaction again after warming the solutions. What changes will julissa most likely see in the warmer reaction.?
A) The bubbling and white solid formation would occur faster
B)The bubbles and white solid formation would happen slower
C) The bubbling or white solid formation would not be created at all
D) Only the bubbling would occur, but no white solid particles would from.
(50 points pls help)
Chemistry
1 answer:
Mamont248 [21]3 years ago
3 0

Answer:

A

Explanation:

With chemical reactions, there are various factors that affect the rate of the reaction. One of these is temperature.

When you raise the temperature, the reaction will move faster. Why? Temperature is directly correlated with the kinetic energy (basically, the energy that makes the particles move). Higher temperatures mean higher kinetic energies. Particles with higher kinetic energies move faster, which makes them more likely to collide. When collisions occur more frequently, the reaction follows through more quickly.

Thus, when Julissa warms the solutions, she will see that bubbling and white solid formation (the products of the reaction) occus faster. So, the answer is A.

Hope this helps!

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The pyrolysis of ethane proceeds with an activation energy of about 300 kJ/mol. How much faster is the decomposition at 625°C th
Alona [7]

Answer:

The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_2 = rate of reaction at T_2

K_1 = rate of reaction at T_1

Ea = activation energy of the reaction

R = gas constant = 8.314 J/K mol

E_a=300 kJ/mol=300,000 J/mol

T_2=625^oC=898.15 K,T_1=525^oC=798.15 K

\log (\frac{K_2}{K_1})=\frac{300,000 J/mol}{2.303\times 8.314 J/K mol}[\frac{1}{798.15 K}-\frac{1}{898.15 K}]

\log (\frac{K_2}{K_1})=2.185666

K_2=153.344\times K_1

The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.

4 0
3 years ago
. a large piece of jewelry has a mass of 132.6 g. a graduated cylinder initially contains 48.6 ml water. when the jewelry is sub
elena-s [515]

The large piece of jewelry  that has a mass of 132.6 g and when is submerged in a graduated cylinder that initially contains 48.6 ml water and the volume increases to 61.2 ml once the piece of jewelry is submerged, has a density of: 10.523 g/ml

To solve this problem the formulas and the procedures that we have to use  are:

  • v = v(f)-v(i)
  • d = m/v

Where:

  • d= density
  • m= mass
  • v= volume
  • v(f) = final volume
  • v(i) = initial volume

Information about the problem:

  • m = 132.6 g
  • v(i) = 48.6 ml
  • v(f) = 61.2 ml
  • v = ?
  • d =?

Applying the volume formula we get:

v = v(f)-v(i)

v = 61.2 ml - 48.6 ml

v = 12.6 ml

Applying the density  formula we get:

d = m/v

d = 132.6 g/12.6 ml

d = 10.523 g/ml

<h3>What is density?</h3>

It is a physical quantity that expresses the ratio of the body mass to the volume it occupies.

Learn more about density in: brainly.com/question/1354972

#SPJ4

3 0
1 year ago
The activation energy of a certain uncatalyzed reaction is 64 kJ/mol. In the presence of a catalyst, the Ea is 55 kJ/mol. How ma
Ksivusya [100]

Answer:

About 5 times faster.

Explanation:

Hello,

In this case, since the Arrhenius equation is considered for both the catalyzed reaction (1) and the uncatalized reaction (2), one determines the relationship between them as follows:

\frac{k_1}{k_2}=\frac{Aexp(-\frac{Ea_1}{RT} )}{Aexp(-\frac{Ea_2}{RT})}  \\\frac{k_1}{k_2}=\frac{exp(-\frac{Ea_1}{RT} )}{exp(-\frac{Ea_2}{RT})}

By replacing the corresponding values we obtain:

\frac{k_1}{k_2}=\frac{exp(-\frac{55000J/mol}{8.314J/molK*673.15K} )}{exp(-\frac{64000J/mol}{8.314J/molK*673.15K} )} =4.8

Such result means that the catalyzed reaction is about five times faster than the uncatalyzed reaction.

Best regards.

4 0
3 years ago
How does velocity and mass affect momentum?
ollegr [7]

Velocity and mass are directly proportional to the quantity of momentum by:

p = mv. Therefore, and increase in either velocity or mass will lead to an increase in momentum and vice versa. Momentum during a reaction is always conserved, meaning that the mass and initial velocity before a reaction will always be equal to the change in mass and velocity produced after the reaction. Kinetic energy after a reaction, however, is not always conserved. For example if a fast moving vehicle collided with a stationary vehicle, and moved together, the overall kinetic energy would be after the reaction, as a heaver mass would be moved by the same velocity causing a decrease in kinetic energy.

I don't know if this is exactly what you are looking for, but in physics this is how it is understood.

4 0
2 years ago
Given the following unbalanced equation:
Lorico [155]
<h3>Answer:</h3>

8.01 mol MgO

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Moles
  • Compounds

<u>Stoichiometry</u>

  • Using Dimensional Analysis
  • Analyzing Reactions RxN
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Unbalanced] Mg + O₂ → MgO

[RxN - Balanced] 2Mg + O₂ → 2MgO

[Given] 8.01 moles Mg

[Solve] moles MgO

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol Mg → 2 mol MgO

<u>Step 3: Stoich</u>

  1. [DA] Set up:                                                                                                       \displaystyle 8.01 \ mol \ Mg(\frac{2 \ mol \ MgO}{2 \ mol \ Mg})
  2. [DA] Multiply/Divide [Cancel out units]:                                                          \displaystyle 8.01 \ mol \ MgO
3 0
2 years ago
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