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yanalaym [24]
2 years ago
5

How many liters of cane juice is needed to supply 5g sucrose if cane juice contains 12% sucrose

Chemistry
1 answer:
pickupchik [31]2 years ago
6 0

Given parameters:

Mass of sucrose  = 5g

Density of sucrose  = 1.12g/mL

Percentage of sucrose per liter of cane juice  = 12%

Unknown:

Volume of cane juice needed = ?

We need to establish the volume - density relationship. Density is the mass of a substance per unit volume.

Mathematically;

                Density  = \frac{mass}{volume}

Now solve for the volume of sucrose;

                  1.12g/mL = \frac{5}{Volume}

       Volume  = \frac{5}{1.12}    =  4.46mL   = 4.46 x 10⁻³L since 1000mL  = 1L

Since 12% of 1 liter of cane juice is sucrose;

                    12% of  x  liter of cane juice  = 4.46 x 10⁻³L

                   Volume of cane juice  = 4.46 x 10⁻³ x \frac{100}{12}   = 0.037L

Volume of cane juice is  0.037L

   

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If the pH of a phosphoric acid (H3PO4) solution is adjusted to 6.50, what is the most abundant species and which is the second m
maria [59]

Explanation:

For the given values of K_{a} we will have the values of pK as follows.

As,        pK_{a} = -log K_{a}

Therefore,

     pK_{a1} = 2.15,     pK_{a2} = 7.20

      pK_{a3} = 12.38

Now, at pH 6.50

      H_{3}PO_{4} \rightarrow H_{2}PO^{-}_{4} + H^{+};   K_{a1}

At pH = 2.15;  H_{3}PO_{4} = H_{2}PO^{-}_{4}

       H_{2}PO^{-}_{4} \rightarrow HPO^{2-}_{4} + H^{+};  K_{a2}

At pH 7.20;  H_{2}PO^{-}_{4} = HPO^{2-}_{4}

        HPO^{2-}_{4} \rightarrow PO^{3-}_{4} + H^{+};     K_{a3}

Hence, we can conclude that most abundant species is H_{2}PO^{-}_{4} and the second most abundant species is HPO^{2-}_{4}.

3 0
2 years ago
6. What is the molarity of 1.35 mol H2So4 in 245<br><br> mL of solution?
vovikov84 [41]

Answer:

5.51mol/L

Explanation:

Number of moles = 1.35moles

Volume of the solution = 245mL = 245*10^-3L = 0.245L

Molarity of a solution is the defined as the number of moles of a solute dissolved in 1L of the solution.

1.35 moles = 0.245L

X moles = 1L

X = (1.35 * 1) / 0.245

X = 5.51mol/L

The molarity of the solution is 5.51mol/L

7 0
1 year ago
Two balloons are charged with an identical quantity and type of charge: 0.004 C. They are held apart at a separation distance of
lukranit [14]

Answer:

The answer to your question is F = 28800 N

Explanation:

Data

q₁ = q₂ = 0.004 C

distance = r = 5 m

k = 9x 10⁹ C

Force = F

Formula

F = k q₁ q₂ / r²

-Substitution

F = (9 x 10⁹)(0.004)(0.004) / (5)²

-Simplification

F = 144000 / 25

-Result

F = 28800 N

-Conclusion

The force of repulsion between these balloons is 28800 N

6 0
2 years ago
A cell was prepared by dipping a Cu wire and a saturated calomel electrode into 0.10 M CuSO4 solution. The Cu wire was attached
Rashid [163]

Answer:

(a)  Cu²⁺ +2e⁻ ⇌ Cu

(c) 0.07 V  

Explanation:

(a) Cu half-reaction

Cu²⁺ + 2e⁻ ⇌ Cu

(c) Cell voltage

The standard reduction potentials for the half-reactions are+

                                             <u> E°/V </u>

Cu²⁺ + 2e⁻ ⇌ Cu;                  0.34  

Hg₂Cl₂ + 2e⁻ ⇌ 2Hg + 2Cl⁻; 0.241

The equation for the cell reaction is

                                                                            E°/V

Cu²⁺(0.1 mol·L⁻¹) + 2e⁻ ⇌ Cu;                               0.34  

<u>2Hg + 2Cl⁻ ⇌ Hg₂Cl₂ + 2e⁻;                             </u> <u>-0.241 </u>

Cu²⁺(0.1 mol·L⁻¹) + 2Hg + 2Cl⁻ ⇌ Cu + Hg₂Cl₂;   0.10

The concentration is not 1 mol·L⁻¹, so we must use the Nernst equation

(ii) Calculations:  

T = 25 + 273.15 = 298.15 K

Q = \dfrac{\text{[Cl}^{-}]^{2}}{ \text{[Cu}^{2+}]} = \dfrac{1}{0.1} = 10\\\\E = 0.10 - \left (\dfrac{8.314 \times 298.15 }{2 \times 96485}\right ) \ln(10)\\\\=0.010 -0.01285 \times 2.3 = 0.10 - 0.03 = \textbf{0.07 V}\\\text{The cell potential is }\large\boxed{\textbf{0.07 V}}

 

3 0
2 years ago
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n200080 [17]
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4 0
2 years ago
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