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8_murik_8 [283]
3 years ago
9

The titration of an acid (H2A) with LiOH solution generates the following titration curve. What are the main components (more th

an half of the initial amount of H2A, besides H2O) at equivalence point 1 (EP1) and equivalence point 2 (EP2)?
Main Components at EP1 Main Components at EP2
A H2A and OH− A2− and OH−
B HA− and Li+ A2− and Li+
C HA− and OH− HA− and Li+
D H2A and Li+ HA− and OH−

Chemistry
1 answer:
faust18 [17]3 years ago
3 0

Answer:

B) HA⁻ and Li⁺ A²⁻ and Li⁺

Explanation:

Titration of acid H₂A with LiOH solution.

At first equivalent point

    H₂A + LiOH → HA⁻ + Li⁺ + H₂O

    Main component at (EP1) => HA⁻ and Li⁺

At second equivalence point

    HA⁻ + LiOH → A²⁻ + Li⁺ + H₂O

    Main component at (EP2) => A²⁻ and Li⁺

Therefore, the correct answer is B

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topjm [15]

Answer:

V = 34.55 L

Explanation:

Given that,

No of moles, n = 1.4

Temperature, T = 20°C = 20 + 273 = 293 K

Pressure, P = 0.974 atm

We need to find the volume of the gas. It can be calculated using Ideal gas equation which is :

PV=nRT

R is gas constant, R=0.08206\ L-atm/mol-K

Finding for V,

V=\dfrac{nRT}{P}\\\\V=\dfrac{1.4\times 0.08206\times 293}{0.974 }\\\\V=34.55\ L

So, the volume of the gas is 34.55 L.

4 0
3 years ago
Round off the following number to 3 digits 34,560​
Vikki [24]

Answer:

3.46x10⁴

Explanation:

Hello,

In this case, we can see that the number 34,560 has five significant figures, it means that if we want to write it with three, we must take the 3, 4 and 5 only. Nevertheless, since the 6 after the five is greater than 5, we can round such five to 6, so we obtain:

346

However, the decimal places cannot get lost, therefore, we move the given thousand  to the three, so the number turns out:

3.46x10⁴

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6 0
3 years ago
A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.
Lubov Fominskaja [6]

Answer:

Ag_2SO_4

Explanation:

Formula for the calculation of no. of Mol is as follows:

mol=\frac{mass\ (g)}{molecular\ mass}

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

Ag, \frac{0.05305}{0.02657} \approx 2

S, \frac{0.02657}{0.02657} \approx 1

O, \frac{0.10594}{0.02657} \approx 4

Therefore, empirical formula of the compound = Ag_2SO_4

7 0
3 years ago
Read 2 more answers
The following thermochemical equation is for the reaction of Fe 3 O 4 (s) with hydrogen (g) to form iron and water vapor Fe 3 O
Mila [183]

Answer:

41.3kJ of heat is absorbed

Explanation:

Based in the reaction:

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<em>1 mole of Fe3O4 reacts with 4 moles of H₂, 151kJ are absorbed.</em>

63.4g of Fe₃O₄ (Molar mass: 231.533g/mol) are:

63.4g Fe₃O₄ × (1mol / 231.533g) = <em>0.274moles of Fe₃O₄</em>

These are the moles of Fe₃O₄ that react. As 1 mole of Fe₃O₄ in reaction absorb 151kJ, 0.274moles absorb:

0.274moles of Fe₃O₄ × (151kJ / 1 mole Fe₃O₄) =

<h3>41.3kJ of heat is absorbed</h3>

<em />

6 0
3 years ago
What is the result at the end of meiosis II?
VMariaS [17]

Answer:

B.) two diploid cells

6 0
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