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Arlecino [84]
3 years ago
11

A chemistry student weighs out of lactic acid into a volumetric flask and dilutes to the mark with distilled water. He plans to

titrate the acid with solution. Calculate the volume of solution the student will need to add to reach the equivalence point. Round your answer to significant digits
Chemistry
1 answer:
34kurt3 years ago
8 0

Answer:

28.0mL of the 0.0500M NaOH solution

Explanation:

<em>0.126g of lactic acid diluted to 250mL. Titrated with 0.0500M NaOH solution.</em>

<em />

The reaction of lactic acid, H₃C-CH(OH)-COOH (Molar mass: 90.08g/mol) with NaOH is:

H₃C-CH(OH)-COOH + NaOH → H₃C-CH(OH)-COO⁻ + Na⁺ + H₂O

<em>Where 1 mole of the acid reacts per mole of the base.</em>

<em />

You must know the student will reach equivalence point when moles of lactic acid = moles NaOH.

the student will titrate the 0.126g of H₃C-CH(OH)-COOH. In moles (Using molar mass) are:

0.126g ₓ (1mol / 90.08g) = <em>1.40x10⁻³ moles of H₃C-CH(OH)-COOH</em>

To reach equivalence point, the student must add 1.40x10⁻³ moles of NaOH. These moles comes from:

1.40x10⁻³ moles of NaOH ₓ (1L / 0.0500moles NaOH) = 0.0280L of the 0.0500M NaOH =

<h3>28.0mL of the 0.0500M NaOH solution</h3>
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Rb has larger radius.

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andreyandreev [35.5K]

Answer: E=∆H*n= -40.6kj

Explanation:

V(CO) =15L=0.015M³

P=11200Pa

T=85C=358.15K

PV=nRT

n=(112000×0.015)/(8.314×358.15)

n(Co)= 0.564mol

V(Co)= 18.5L = 0.0185m³

P=744torr=98191.84Pa

T= 75C = 388.15k

PV=nRT

n= (99191.84×0.0185)/(8.314×348.15)

n(H2) = 0.634mol

n(CH30H) =1/2n(H2)=1/2×0.634mol

=0.317mol

∆H =∆Hf{CH3OH}-∆Hf(Co)

∆H=-238.6-(-110.5)

∆H = 128.1kj

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3 0
3 years ago
0.475 g H, 7.557 gS, 15.107 g O. Express your answer as a chemical formula.
max2010maxim [7]

Answer:

H₂SO₄

Explanation:

We have a compound formed by 0.475 g H, 7.557 g S, 15.107 g O. In order to determine the empirical formula, we have to follow a series of steps.

Step 1: Calculate the total mass of the compound

Total mass = mass H + mass S + mass O = 0.475 g + 7.557 g + 15.107 g

Total mass = 23.139 g

Step 2: Determine the percent composition.

H: (0.475g/23.139g) × 100% = 2.05%

S: (7.557g/23.139g) × 100% = 32.66%

O: (15.107g/23.139g) × 100% = 65.29%

Step 3: Divide each percentage by the atomic mass of the element

H: 2.05/1.01 = 2.03

S: 32.66/32.07 = 1.018

O: 65.29/16.00 = 4.081

Step 4: Divide all the numbers by the smallest one

H: 2.03/1.018 ≈ 2

S: 1.018/1.018 = 1

O: 4.081/1.018 ≈ 4

The empirical formula of the compound is H₂SO₄.

7 0
3 years ago
Determine the concentration of a solution of sodium hydroxide (NaOH) using a 0.500M Potassium Hydrogen Phthalate (KHP) standard
Svetradugi [14.3K]

Answer:

0.425M NaOH assuming the volume of KHP was 25.50mL and the volume of the NaOH solution was 30.0mL

Explanation:

The KHP reacts with NaOH as follows:

KHP + NaOH → KNaP + H₂O

<em>Where 1 mole of KHP reacts per mole of KNaP</em>

<em />

That means, the moles of KHP added to the NaOH solution = Moles NaOH at equivalence point. With the moles of NaOH and the volume in liters we can find the molar concentration of NaOH.

<em>Assuming the volume added of KHP was 25.50mL and the solution of NaOH contains 30.0mL (0.0300L), the concentration of the NaOH is:</em>

<em />

<em>Moles KHP = Moles NaOH:</em>

25.50mL = 0.02550L * (0.500mol / L) = 0.01275 moles KHP = Moles NaOH

<em>Molarity NaOH:</em>

0.01275 moles NaOH / 0.0300L =

<h3>0.425M NaOH assuming the volume of KHP was 25.50mL and the volume of the NaOH solution was 30.0mL</h3>
7 0
3 years ago
Can someone help me? It needs to have a diagram that has arrows.
daser333 [38]

Answer: The enthalpy change for formation of butane is -125 kJ/mol

Explanation:

The balanced chemical reaction is,

C_4H_{10}(g)+\frac{13}{2}O_2(g)\rightarrow 4CO_2(g)+5H_2O(l)

The expression for enthalpy change is,

\Delta H=[n\times H_f{products}]-[n\times H_f{reactants}]

Putting the values we get :

\Delta H=[4\times H_f_{CO_2}+5\times H_f_{H_2O}]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times H_f_{O_2}]

-2877=[(4\times -393)+(5\times -286)]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times 0]

H_f_{C_4H_{10}=-125kJ/mol

Thus enthalpy change for formation of butane is -125 kJ/mol

5 0
3 years ago
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