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UNO [17]
3 years ago
13

Assuming the densities are the same as water, calculate the concentration in millimolar (mM) for all three reactant solutions, t

hen (assuming one drop weighs 20 mg) calculate the millimoles of each one you used and the mole ratios, all relative to acrylic acid. Include the mole ratio of sodium hydroxide. What percent of the acrylic acid groups were neutralized
Chemistry
1 answer:
Marina86 [1]3 years ago
7 0

Answer:

foatation

Explanation:

You might be interested in
Choose your favorite super hero that gained their powers from radiation. What
dmitriy555 [2]

Answer:

Copy and paste "Electromagnetic waves are categorized according to their frequency f or, equivalently, according to their wavelength λ = c/f.  Visible light has a wavelength range from ~400 nm to ~700 nm.  Violet light has a wavelength of ~400 nm, and a frequency of ~7.5*1014 Hz.  Red light has a wavelength of ~700 nm, and a frequency of ~4.3*1014 Hz." into google, and the correct website pops up as the first result.

Explanation:

I tried to link the website that I use to convert wavelengths and frequencies into types of light, but it deleted my answer, so I guess we're doing it this way. As for converting the wavelength to energy, the same principles apply as before:

Frequency: ν       Wavelength: λ       Energy: E       Speed of light: C (3.00e8)       Planck's Constant: h (6.626e-34)

ν -> λ    λ = C/ν

λ -> ν    ν = C/λ

For either of these equations, wavelength must be converted to meters or nanometers, depending on the equation.

For ν -> λ, after doing the equation, convert the wavelength into nanometers by dividing by 1e-9.

For converting λ -> ν, convert the wavelength into meters by multiplying by 1e-9.

For energy: E = hν = hc/λ

6 0
3 years ago
If a 95.27 mL sample of acetic acid (HC2H3O2) is titrated to the equivalence point with 79.06 mL of 0.113 M KOH, what is the pH
KATRIN_1 [288]

Answer:

8.73

Explanation:

The concentration of acetic acid can be determined as follows:

M_1V_1 = M_2V_2\\(KOH) = (CH_3COOH)

M_{KOH}=0.113 M\\V_{KOH}=79.06 mL

V_{CH_3COOH}=95.27 \\\\M_{CH_3COOH)=?????

M_{CH3COOH} = \frac{M_{KOH}*V_{KOH}}{V_{CH_3COOH}}

M_{CH3COOH} = \frac{0.113*79.06}{95.27}

M_{CH3COOH} = 0.094 M

Moles of CH_3COOH = 95.27* 10^{-3}* 0.094

=0.0090 moles

Moles of  KOH = 79.06*10^{-3}*0.113

= 0.0090 moles

The equation for the reaction can be expressed as :

CH_3COOH     +      KOH     ----->      CH_3COO^{-}K^+      +     H_2O

Concentration of CH_3COO^{- ion = \frac{0.0090}{Total volume (L)}

= \frac{0.0090}{(95.27+79.06)} *1000

= 0.052 M

Hydrolysis of  CH_3COO^{- ion:

CH_3COO^{-      +       H_2O      ----->      CH_3COOH       +     OH^-

K = \frac{K_w}{K_a} = \frac{x*x}{0.052-x}

⇒    \frac{10^{-14}}{1.82*10^{-5}}= \frac{x*x}{0.052-x}

=     0.5494*10^{-9}= \frac{x*x}{0.052-x}

As K is so less, then x appears to be a very infinitesimal small number

0.052-x ≅ x

0.5494*10^{-9}= \frac{x^2}{0.052}

x^2 = 0.5494*10^{-9}*0.052

x^2 = 0.286*10^{-10

x = \sqrt{0.286*10^{-10

x =0.535*10^{-5}M

[OH] = x =0.535*10^{-5}

pOH = -log[OH^-]

pOH = -log[0.535*10^{-5}]

pOH = 5.27

pH = 14 - pOH

pH = 14 - 5.27

pH = 8.73

Hence, the pH of the titration mixture = 8.73

8 0
3 years ago
What is the organelle that packages and releases materials other parts of the cell?
VARVARA [1.3K]
The vacuoles are the answer
7 0
4 years ago
Which of the following are true statements about equilibrium systems?For the following reaction at equilibrium:2 H2(g) + O2(g) ?
VMariaS [17]

These are five questions about equilibrium systems each with its complete answer.

<u>Question 1</u><u>.</u> For the following reaction at equilibrium:

2 H₂(g) + O₂(g) ⇄ 2 H2O(g),  the equilibrium will shift to the left if the volume is doubled?

Answer: TRUE

Explanation:

When a force disturbs a chemical <em>equillibrium</em>, the system will shift toward the direction that <em>reduces the effect</em>. This is Le Chatelier's principle.

As per Bolye's law, at constant temperature, the volume and the pressure of a fixed amount of gas are inversely related.

Also, the pressure of the system is directly related to the number of particles (atoms or molecules). Hence, more molecules, more pressure; less molecules, less pressure.

Now, you can reason in this way: if the volume of the given system is doubled, then the pressure is lowered, and the system will try to alleviate this disturbance by shifting the reaction to the side that produces more molecules, to restore the pressure.  Because on the left side three molecules can be produced from the reaction of two molecules of H₂O on the rihgt, <em>the system will shift to the left</em>. And this proves the truth of the statement.

<u>Question 2</u>. For the following reaction at equilibrium:

H₂(g) + F₂(g) ⇄  2HF(g), removing H₂ will decrease the amount of F₂ present once equilibrium is reestablished.

Answer: FALSE.

Explanation:

Note that, since the temperature and other conditions have not changed, the equilibrium constant, Ke, has not changed. And, for the given equilibrium, Ke is given by the following equation.

  • Ke = [ H₂] [F₂] / [HF]²

Hence, to keep Ke unchanged, when removing H₂, the amount of F₂ present once equilibrium is reestablished will have to increase.

This is the opposite of the stated on the question, so the statement is false.

<u>Question 3.</u> Increasing the temperature of an exothermic reaction shifts the equilibrium position to the right.

Answer: FALSE.

Explanation:

You can write an <em>exothermic equlibrium</em> placing heat as a product on the right side of the equation; in this way:

  • A + B ⇄ C + D + heat

There, treating the heat as another product, you can reason that increasing the temperature, which is equivalent to supplying heat, will shift the equilibrium to the left side to consume heat, instead to the proposed by the statement. So, this is a false statement.

<u>Question 4</u>. For the following reaction at equilibrium:

CaCO₃(s) ⇄ CaO(s) + CO₂ (g), adding more CaCO₃ will shift the equilibrium to the right.

Answer: TRUE.

Explanation:

CaCO₃(g) is the only reactant of the forward reaction.

Adding more CaCO₃ may be seen as a disturbance against which the system will act by consuming it and producing more CaO and CO₂.

So, the forward reation will be favored and you conclude that <em>adding more CaCO₃ will shift the equilibrium to the right.</em>

<u>Question 5.</u> For the following reaction at equilibrium:

CaCO₃(s) ⇄ CaO(s) + CO₂ (g), increasing the total pressure by adding Ar(g) will have no effect on the equilibrium position.

Answer: TRUE.

Explanation:

In accordance to Le Chatelier's principle, increasing the pressure should be addresed by the equilibrium by shifting to the side where such pressure increase could be released.

That is possible when the number of molecules of gases on both sides are different: the equilibrium will shift to the side where more molecules less molecules are produced.

But, when the stoichiometry of the reaction shows the same number of molecules on both sides, which is the case in the given equilibrium, increasiing (or decreasing) the pressure will have no effect on the equilibrium position. Then, the answer is true.

8 0
3 years ago
A reaction contains two reactants, A and B. If A is doubled, there will be a greater number of effective collisions between reac
liubo4ka [24]

Answer:

True

Explanation:

The complete question is:

<u><em>"A reaction contains two reactants, A and B. If A is doubled, there will be a greater number of effective collisions between reactants. TRUE FALSE"</em></u>

Collision Theory indicates that chemical reactions take place because molecules, atoms or ions collide with each other.

Furthermore, the molecules must collide effectively, that is, not all reagent collisions lead to product formation. Effective shock means that the reagent molecules have enough kinetic energy at the time of the shock for their bonds to break and product bonds to form. In addition, the molecules of the reagents must be properly oriented for the reaction to take place.

As the concentration increases, the number of shocks increases. In other words, by increasing the concentration of the reactants, the probability of collision between their molecules increases, and therefore the number of effective collisions.So the statement is true-

6 0
3 years ago
Read 2 more answers
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