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icang [17]
2 years ago
12

Alka‑Seltzer is marketed as a remedy for stomach problems, such as heartburn or indigestion, and pain relief. It contains aspiri

n, sodium bicarbonate, and citric acid, and is effervescent when placed in water. Enter the equation for the reaction of one sodium bicarbonate ( NaHCO3 ) molecule with one citric acid ( C 6 H 8 O 7 ) molecule. Do not include phases.
Chemistry
1 answer:
Papessa [141]2 years ago
5 0

Answer:

The equation for the reaction of one sodium bicarbonate ( NaHCO3 ) molecule with one citric acid (C6H8O7) molecule is the following:

Sodium Bicarbonate + Citric Acid ⇒ Water + Carbon Dioxide + Sodium Citrate

NaHCO3 + C6H8O7 ⇒ 3 CO2 + 3 H2O + Na3C6H5O7

Explanation:

The reaction is in balance, that is, the whole H2CO3 is not finished, but a little bit of this acid is left in the solution. Therefore, when sodium bicarbonate is added to the solution with citric acid, sodium citrate salt (C6H5O7Na3) and carbonic acid (H2CO3) are formed, which is rapidly broken down into water (H2O) and carbonic oxide (CO2).

C6H8O7 + NaHCO3 ⇒ C6H5O7Na3 + 3 H2CO3

C6H5O7Na3 + 3 H2CO3 ⇔ C6H5O7Na3 + 3 H2O + 3 CO2

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The first five ionization energies of an element are as follows (in kJ/mol): 577.9, 1820, 2750, 11600, 14800. Which of the follo
Ne4ueva [31]

Answer:option d==> Si.

Explanation:

The energy required to remove electron from a gaseous atom or ion is what is called an ionization energy. As we remove electrons continually in a gaseous atom or ion, the ionization energy increases which are know as the first ionization energy, the second ionization energy, third ionization energy and so on.

Looking at the electronic configuration of Silicon, Si; Ne 3s2 3p2. We can can see that the first four ionization energies are from the removal of the 3p2 and 3s2 electrons and the fifth ionization energy, which is the highest ionization energy of 14800 kJ/mol is the the electron removed from the core shell.

4 0
2 years ago
Question 096 Propose a three-step synthetic sequence to accomplish the transformation below. 1) HBr, ROOR; 2) t-BuOK; 3) CH3CH2C
salantis [7]

Complete Question

Question 096 Propose a three-step synthetic sequence to accomplish the transformation below.

Option 1 =>  1) HBr, ROOR; 2) t-BuOK; 3) CH3CH2CCNa

Option 2 => 1) NaOEt; 2) HBr, ROOR; 3) CH3CH2CCNa

Option 3 => 1) t-BuOK; 2) NaNH2; 3) CH3CH2CCNa

Option 4 => 1) CH3CH2CH2CH2Br; 2) NaOEt; 3) HBr, ROOR

Option 5 => 1) t-BuOK; 2) HBr, ROOR; 3) CH3CH2CCNa

Option 6 => 1) NaOEt; 2) NBS, hν; 3) NaSBu

Answer:

The correct option is option 5

Explanation:

   The mechanism of the reaction is shown on the first uploaded image

8 0
2 years ago
A student pours exactly 26.9 mL of HCl acid of unknown molarity into a beaker. The student then adds 2 drops of the indicator an
Assoli18 [71]
a.
Acids react with bases and give salt and water and the products.

Hence, HCl reacts with NaOH and gives NaCl salt and H₂O as the products. The reaction is,
            HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

To balance the reaction equation, both sides hould have same number of elements.

Left hand side,                                             Right hand side,
             
H atoms = 2                                               H atoms = 2
            Cl atoms = 1                                               Cl atoms = 1
            Na atoms = 1                                               Na atoms = 1 
           O atoms = 1                                                   O atoms = 1

Hence, the reaction equation is already balanced.

b. 
Molarity (M)= moles of solute (mol) / Volume of the solution (L)
 
          HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Molarity of NaOH = <span>0.13 M
</span>Volume of NaOH added = <span>43.7 mL
Hence, moles of NaOH added = 0.13 M x 43.7 x 10</span>⁻³ L
                                                 = 5.681 x 10⁻³ mol

Stoichiometric ratio between NaOH and HCl is 1 : 1

Hence, moles of HCl = moles of NaOH
                                    = 
5.681 x 10⁻³ mol

5.681 x 10⁻³ mol of HCl was in <span>26.9 mL.

Hence, molarity of HCl = </span>5.681 x 10⁻³ mol / 26.9 x 10⁻³ L
                                     = 0.21 M
6 0
2 years ago
The nucleoside adenosine exists in a protonated form with a pKa of 3.8. The percentage of the protonated form at pH 4.8 is close
Natasha_Volkova [10]

Answer:

Ok:

Explanation:

So, you can use the Henderson-Hasselbalch equation for this:

pH = pKa + log(A^-/HA) where A- is the conjugate base of the acid. In other words, A- is the deprotonated form and HA is the protonated.

We can solve that

1 = log(A^-/HA\\) and so 10 = A^-/HA or 10HA = A-.  For every 1 protonated form of adenosine (HA), there are 10 A-. So, the percent in the protonated form will be 1(1+10) or 1/11 which is close to 9 percent.

6 0
2 years ago
At 800 K, the equilibrium constant, Kp, for the following reaction is 3.2 × 10–7. 2 H2S(g) ⇌ 2 H2(g) + S2(g) A reaction vessel a
djverab [1.8K]

Answer:

0.008945 atm

Explanation:

In the reaction:

2H2S(g) ⇌ 2 H2(g) + S2(g)

Kp is defined as:

Kp = P_{H_{2}}^2P_{S_{2}} / P_{H_{2}S}^2

<em>Where P is the pressure of each compound in equilibrium.</em>

If initial pressure of H2S is 3.00atm, concentrations in equilibrium are:

H2S = 3.00 atm - 2X

H2 = 2X

S2: = X

Replacing:

3.2x10^{-7} = (2X)^2X / (3-2X)^2

3.2x10^{-7} = 4X^3 / 9- 6X+4X^2

0 = 4X³ - 1.28x10⁻⁶X² + 1.92x10⁻⁶X - 2.88x10⁻⁶

Solving for X:

X = 0.008945 atm

As in equilibrium, pressure of S2 is X, <em>pressure is 0.008945 atm</em>

7 0
2 years ago
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