What is the identity of this atom?

CH3CH2OH + 302 → 2002 + 3H20

Softa [21]

Answer:

Oxygen gas

Explanation:

Given expression:

CH₃CH₂OH + 3O₂ → 2CO₂ + 3H₂O

Number of moles of oxygen gas = 1 mole

Number of moles of ethanol = 1 mole

Unknown:

The limiting reactant = ?

Solution:

The limiting reactant is the reactant in short supply in a chemical reaction. To find this specie, we always us the number of moles.

From the equation of the reaction:

1 mole of ethanol would require 3 moles of oxygen gas

But we have been given just 1 mole of oxygen gas instead of 3moles.

Therefore, oxygen gas is the limiting reactant.

3 0

3 years ago

Amphetamine (C9H13N)(C9H13N) is a weak base with a pKbpKb of 4.2. You may want to reference (Pages 710 - 713) Section 16.8 while

Soloha48 [4]

Answer:

pH = 10.38

Explanation:

C9H13N ↔ C9H20O3N+ + OH-

∴ molar mass C9H13N = 135.21 g/mol

∴ pKb = - log Kb = 4.2

⇒ Kb = 6.309 E-5 = [OH-][C9H20O3N+] / [C9H13N]

∴ C sln = (205 mg/L )*(g/1000 mg)*(mol/135.21 g) = 1.516 E-3 M

mass balance:

⇒ C sln = 1.516 E-3 = [C9H20O3N+] + [C9H13N]......(1)

charge balance:

⇒ [C9H20O3N+] + [H3O+] = [OH-]; [H3O+] is neglected, come from water

⇒ [C9H20O3N+] = [OH-].......(2)

(2) in (1):

⇒ [C9H13N] = 1.516 E-3 - [OH-]

replacing in Kb:

⇒ Kb = 6.3096 E-5 = [OH-]² / (1.516 E-3 - [OH-])

⇒ [OH-]² + 6.3096 E-5[OH] - 7.26613 E-8 = 0

⇒ [OH-] = 2.3985 E-4 M

∴ pOH = - Log [OH-]

⇒ pOH = 3.62

⇒ pH = 14 - pOH = 14 - 3.62 = 10.38

5 0

3 years ago

g. What is crystallization?Name any two go mixtures that can be separated by this method:name any two mixtures that can be separ

Masteriza [31]

crystallization is a process which helps to separate a pure solid from a solution in its crystal form. This is the in use to purify solid. For an example the salt we get from seawater can have many impurities in it. Hence, the process of crystallization is in use to remove these impurities.

Sugar and salt are examples of products where crystallization does not only serve as separation/purification technique, but where it is also responsible for getting crystals with the right size (and shape) for further application of the products.

- BRAINLIEST answerer

7 0

2 years ago

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The decomposition of \rm XY is second order in \rm XY and has a rate constant of6.96Ã10â3M^{-1} \cdot s^{-1} at a certain temper

LUCKY_DIMON [66]

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Half life for second order kinetics is given by:

$t_{1/2}=\frac{1}{k\times a_0}$

k = rate constant =?

$a_0$ = initial concentration

a = concentration left after time t

Integrated rate law for second order kinetics is given by:

$\frac{1}{a}=kt+\frac{1}{a_0}$

a) Initial concentration of XY = $a_o=0.100 M$

Rate constant of the reaction = k = $6.96\times 10^{-3} M^{-1} s^{-1}$

Half life of the reaction is:

$t_{1/2}=\frac{1}{6.96\times 10^{-3} M^{-1} s^{-1}\times 0.100 M}$

$=1,436.78 s$

1,436.78 seconds is the half-life for this reaction.

b) Initial concentration of XY = 0.100 M

Final concentration after time t = 12.5% of 0.100 M = 0.0125 M

$\frac{1}{0.0125 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.100M}$

Solving for t;

t = 10,057.47 seconds

In 10,057.47 seconds the concentration of XY will become 12.5% of its initial concentration.

c) Initial concentration of XY = 0.200 M

Final concentration after time t = 12.5% of 0.200 M = 0.025 M

$\frac{1}{0.025 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}$

Solving for t;

t = 5,028.73 seconds

In 5,028.73 seconds the concentration of XY will become 12.5% of its initial concentration.

d) Initial concentration of XY = 0.160 M

Final concentration after time t = $6.20\times 10^{-2} M$

$\frac{1}{6.20\times 10^{-2} M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}$

Solving for t;

t = 1,419.40 seconds

In 1,419.40 seconds the concentration of XY will become $6.20\times 10^{-2} M$ .

e) Initial concentration of $SO_2Cl_2= 0.050 M$

Final concentration after time t = x

t = 55.0 s

$\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 55.0 s+\frac{1}{0.050 M}$

Solving for x;

x = 0.04906 M

The concentration after 55.0 seconds is 0.04906 M.

f) Initial concentration of XY= 0.050 M

Final concentration after time t = x

t = 500 s

$\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 500 s+\frac{1}{0.050 M}$

Solving for x;

x = 0.04259 M

The concentration after 500 seconds is 0.0.04259 M.

7 0

3 years ago

A student does not observe a change when holding a test tube in a flame. However, a change is expected. What is the most likely

dalvyx [7]

The answer is: The reactants were not heated long enough.

For all chemical reaction some energy is required and that energy is called activation energy (energy that needs to be absorbed for a chemical reaction to start).

Activation energy is the minimum energy colliding particles must have in order to react.

By lowering activation energy, reaction need less heat.

In this example, there is not enough heat.

3 0

3 years ago

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