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abruzzese [7]
2 years ago
12

A chemist reacts sodium metal (Na) and chlorine gas (Cl) to form salt according to the balanced chemical equation: Na + Cl → NaC

l. If the chemist has 35.0 g Na, what mass of chlorine must he or she use to react completely with the sodium? Finally, solve (remember significant figures)
What is A?

Couldn't wait so I got it wrong but for all the people needing this question answered the answer is A= 54.1

Chemistry
2 answers:
Brums [2.3K]2 years ago
4 0

The reaction involves adding Na and Cl₂ gas to form NaCl

since chlorine is a halogen it exists as a diatomic molecule - Cl₂

the balanced equation is

2Na + Cl₂ --> 2NaCl

the molar ratio reactants Na to Cl₂ is 2:1

the mass of Na present is 35.0 g

therefore number of moles of Na is - 35.0 g / 23 g/mol = 1.52 mol

according to the molar ratio , the number of Cl₂ moles required is half the number of Na moles present

If 2 moles of Na reacts with 1 mol of Cl₂

then 1.52 mol of Na reacts with - 1/2 x 1.52 mol = 0.760 mol of Cl₂

molar mass of Cl₂ is - 71.0 g/mol

number of Cl₂ moles are - 0.760 mol x 71.0 g/mol = 54.0 g

answer is 54.0 g of Cl₂ is needed

EastWind [94]2 years ago
4 0

54.1g Cl is the answer.                                        

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The pH of an aqueous solution is 4.32. What is the [OH–]?
Molodets [167]
2.10 x 10^-10 M. Ans


pH + pOH = 14
Where, pOH is the power of hydroxide ion concentration and pH is the power of concetration of the H+ ion.
Now, pOH = 14 - 4.32
= 9.68
Now, the concentration of [H+] is 10-7 M, then pH is 7 and for [OH-] = 10-7 M, the pOH is also 7.

Now, pOH = -log[OH-]
[OH-] = 10^- pOH
= 10^-9.68
= 2.10 x 10^-10 M

8 0
2 years ago
Using your data above, draw conclusions about the d-splitting for each ligand (H2O, en, phen). Order the complexes from least to
Delvig [45]

Answer:

H2O<en<phen

Explanation:

The degree of d- splitting is observed from the intensity of colour. The order of d splitting from least to greatest is H2O<en<phen. Phen shows the greatest d-splitting. The degree of splitting of d- orbitals by ligands depends on their relative positions in the spectrochemical series. The spectrochemical series is an experimentally determined series. The series separates the ligands into strong field and weak field ligands. Strong field ligands are found towards the end of the series. Strong field ligands such as en and phen can participate in metal to ligand or ligand to metal pi-bonding. Hence they cause more d-splitting. Ethylendiamine and phenanthroline occur towards the end of the spectrochemical series hence the higher order of d-splitting.

7 0
2 years ago
trans-2-Butene does not exhibit a signal in the double-bond region of the spectrum (1600–1850 cm-1); however, IR spectroscopy is
spayn [35]

Answer:

The other signal that would indicate the presence of a C= C bond appears close to 3100 cm^{-1}.

Explanation:

Bands that appear above 3000 cm^{-1}  are often unsaturation diagnoses suggest. The band at 3000- 3100 cm^{-1} is characteristics for C-H stretching frequencies and normally is overlaps with the ones for alkanes because it is a band of weak intensity.

4 0
2 years ago
The beta oxidation pathway degrades activated fatty acids (acyl-CoA) to acetyl-CoA, which then enters the citric acid cycle. Add
BabaBlast [244]

Answer:

The correct statements are given below

Explanation:

b Enoyl CoA isomerase an enzyme that converts cis double bonds to trans double bonds in fatty acid metabolism,bypasses a step that reduces Q,resulting in the higher ATP yield.

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3 0
2 years ago
Read 2 more answers
If 10.0 grams of NaHCO3 is added to 10.0 g of HCl, determine the efficiency of baking soda as an antacid if 6.73 g of NaCl was p
Lapatulllka [165]

Answer:

percentage yield of NaCl = 96.64%

Explanation:

The reaction was between NaHCO3 and HCl .The chemical equation can be represented below:

NaHCO3 + HCl → NaCl + H2O + CO2 . The balance equation is

NaHCO3 + HCl → NaCl + H2O + CO2

The question ask us to calculate the percentage yield of NaCl.

The efficiency of NaHCO3 as an antacid , the limiting reactant is NaHCO3

as

1 mole of NaHCO3 produces 1 mole of NaCl

Therefore,

molar mass of NaHCO3 = 23 +1 + 12 + 48 = 84 g

molar mass of NaCl = 23 + 35.5 = 58.5 g

1 mole of NaHCO3 = 84 g

1 mole of NaCl  = 58.5 g

since 84 g of NaHCO3 produces 58.5 g of NaCl

10 g of NaHCO3 will produce ? grams of NaCl

cross multiply

Theoretical yield of NaCl = (10 × 58.5)/84

Theoretical yield of NaCl = 585/84

Theoretical yield of NaCl  = 6.9642857143 g

percentage yield of NaCl = actual yield/theoretical yield × 100

percentage yield of NaCl = 6.73/6.9642857143 × 100

percentage yield of NaCl = 673/6.9642857143

percentage yield of NaCl = 96.635897436%

percentage yield of NaCl = 96.64%

3 0
2 years ago
Read 2 more answers
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