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nirvana33 [79]
3 years ago
8

Please help :/ need this done in 15 minutes mannn :(

Chemistry
1 answer:
IgorLugansk [536]3 years ago
4 0

Answer:hey man am sorry but have you tried to search it up

Explanation:

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g If 50.0 mL of a 0.75 M acetic acid solution is titrated with 1.0 M sodium hydroxide, what is the pH after 10.0 mL of NaOH have
V125BC [204]

Answer:

pH = 2.66

Explanation:

  • Acetic Acid + NaOH → Sodium Acetate + H₂O

First we <u>calculate the number of moles of each reactant</u>, using the <em>given volumes and concentrations</em>:

  • 0.75 M Acetic acid * 50.0 mL = 37.5 mmol acetic acid
  • 1.0 M NaOH * 10.0 mL = 10 mmol NaOH

We<u> calculate how many acetic acid moles remain after the reaction</u>:

  • 37.5 mmol - 10 mmol = 27.5 mmol acetic acid

We now <u>calculate the molar concentration of acetic acid after the reaction</u>:

27.5 mmol / (50.0 mL + 10.0 mL) = 0.458 M

Then we <u>calculate [H⁺]</u>, using the<em> following formula for weak acid solutions</em>:

  • [H⁺] = \sqrt{C*Ka}=\sqrt{0.458M*1.76x10^{-5}}
  • [H⁺] = 0.0028

Finally we <u>calculate the pH</u>:

  • pH = -log[H⁺]
  • pH = 2.66
8 0
3 years ago
One way of obtaining pure sodium carbonate is through the decomposition of the mineral trona, Na3(CO3)(HCO3)·2H2O. 2Na3(CO3)(HCO
zhenek [66]
Percentage yield = (actual yield / theoretical yield) x 100%

The balanced equation for the decomposition is,
 2Na₃(CO₃)(HCO₃)·2H₂O(s) → 3Na₂CO₃(s) + CO₂(g) + 5H₂<span>O(g)

The stoichiometric ratio between </span>Na₃(CO₃)(HCO₃)·2H₂O(s)  and Na₂CO₃(s) is 2 : 3

The decomposed mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 1000 kg
                                                                                     = 1000 x 10³ g

Molar mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 226 g mol⁻¹
moles of Na₃(CO₃)(HCO₃)·2H₂O(s) = mass / molar mass
                                                         = 1000 x 10³ g / 226 g mol⁻¹
                                                         = 4424.78 mol

Hence, moles of Na₂CO₃ formed = 4424.78 mol x \frac{3}{2}
                                                     = 6637.17 mol

Molar mass of Na₂CO₃ = 106 g mol⁻¹

Hence, mass of Na₂CO₃ = 6637.17 mol x 106 g mol⁻¹
                                        = 703540.02 g
                                        = 703.540 kg

Hence, the theoretical yield of Na₂CO₃ =  703.540 kg
Actual yield of Na₂CO₃ = 650 kg

Percentage yield = (650 kg / 703.540 kg) x 100%
                            = 92.34%
7 0
3 years ago
Which element most likely has the same number of valence electrons as silicon(si)?
Blizzard [7]

Answer:

Carbon, germanium, tin and lead.

Explanation:

The silicon is belong to the carbon family. There are five elements in carbon family carbon, silicon, germanium, tin and lead. These five elements are present in same group i.e group fourteen. The elements present in same group have same number of valance electrons.

For example.

Carbon electronic configuration:

C₆ = [He]  2s² 2p²

Silicon electronic configuration:

Si₁₄ = [Ne] 3s² 3p²

Germanium electronic configuration:

Ge₃₂ = [Ar] 3d¹⁰ 4s² 4p²

Tin electronic configuration:

Sn₅₀ = [Kr] 4d¹⁰ 5s² 5p²

Lead electronic configuration:

Pb₈₂ = [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p²

we can see that in case of all elements there are four valance electrons, which are equal to the valance electrons of silicon.

7 0
3 years ago
Which of the following is an example of a chemical change? *
Norma-Jean [14]

Answer:

hydrogen and oxygen to form water

3 0
3 years ago
Which of the following would be found in both eukaryotic AND prokaryotic cells?
beks73 [17]
Nucleus ,endoplasmic reticulum
7 0
3 years ago
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