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3 years ago

Would a reaction occur if a ZnSo4 solution and a SnSO4 solution were mixed?

Chemistry

1 answer:

Mrrafil [7]3 years ago

8 0

Hey there!

No !! there is nothing to react. What else other than tin sulphate and zinc sulphate can you make - but then you have only what you started with .

Hope that helps!

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Calculate the number of repetitions of the β-oxidation pathway required to fully convert a 18-carbon activated fatty acid to ace

Anna35 [415]

The number of repetition of the Beta oxidation requires to fully convert 18 carbon activated fatty acid to Acetyl COA molecule is NINE.
Beta oxidation of fatty acid involves oxidative removal of successive two carbon units in the form of acetlyl COA, starting from the carboxyl end of the fatty acyl chain. Thus, a fatty acid which has 18 carbons will have to undergo nine repetition of beta oxidation before it is reduced to an acetyl COA.

4 0

3 years ago

How many miles of iron can be produced from 3.5 moles of Fe2O3 and 6.2 moles of CO

AlekseyPX

Answer:

4.13 moles of Fe.

Explanation:

Given data:

Moles of iron produced = ?

Moles of Fe₂O₃ = 3.5 mol

Moles of CO = 6.2 mol

Solution:

Chemical equation:

Fe₂O₃ + 3CO → 2Fe + 3CO₂

Now we will compare the moles of iron with CO and Fe₂O₃.

Fe₂O₃ : Fe

1 : 2

3.5 : 2/1×3.5 = 7 mol

CO : Fe

3 : 2

6.2 : 2/3×6.2 = 4.13 mol

The number of moles of iron produced by CO are less it will limiting reactant.

Thus, moles of iron formed in given reaction are 4.13 moles.

6 0

3 years ago

Using the thermodynamic information in the ALEKS Data tab, calculate the boiling point of titanium tetrachloride . Round your an

ddd [48]

Answer:

The boiling point is 308.27 K (35.27°C)

Explanation:

The chemical reaction for the boiling of titanium tetrachloride is shown below:

Ti $Cl_{4(l)}$ ⇒ Ti $Cl_{4(g)}$

ΔH° $_{f}$ (Ti $Cl_{4(l)}$ ) = -804.2 kJ/mol

ΔH° $_{f}$ (Ti $Cl_{4(g)}$ ) = -763.2 kJ/mol

Therefore,

ΔH° $_{f}$ = ΔH° $_{f}$ (Ti $Cl_{4(g)}$ ) - ΔH° $_{f}$ (Ti $Cl_{4(l)}$ ) = -763.2 - (-804.2) = 41 kJ/mol = 41000 J/mol

Similarly,

s°(Ti $Cl_{4(l)}$ ) = 221.9 J/(mol*K)

s°(Ti $Cl_{4(g)}$ ) = 354.9 J/(mol*K)

Therefore,

s° = s° (Ti $Cl_{4(g)}$ ) - s°(Ti $Cl_{4(l)}$ ) = 354.9 - 221.9 = 133 J/(mol*K)

Thus, T = ΔH° $_{f}$ /s° = [41000 J/mol]/[133 J/(mol*K)] = 308. 27 K or 35.27°C

Therefore, the boiling point of titanium tetrachloride is 308.27 K or 35.27°C.

5 0

3 years ago

(LO 3N, 4G, 4O) Aluminum sulfate is also involved in dying fabrics. The gelatinous precipitate formed in the reaction with dilut

brilliants [131]

9.4 × 10⁻³ mg (0.73 mmoles) of Al(OH)₃ is formed

Explanation:

We have the following chemical reaction:

Al₂(SO₄)₃(aq) + 6 NaOH(aq) → 2 Al(OH)₃(s) + 3 Na₂SO₄(aq)

The precipitate mentioned by the problem is aluminium hydroxide Al(OH)₃.

Now to determine the number of moles of sodium hydroxide NaOH we use the following formula:

molar concentration = number of moles / volume

number of moles = molar concentration × volume

number of moles of NaOH = 0.088 M × 25 mL = 2.2 mmoles

number of moles of Al₂(SO₄)₃ = 5.6 × 10⁻³ moles = 5.6 mmoles (found in the problem text)

We see from the chemical reaction that 1 mole of Al₂(SO₄)₃ requires 6 moles of NaOH so 5.6 mmoles of Al₂(SO₄)₃ would require 6 times more NaOH which is 33.6 mmoles and we have only 2.2 mmoles. The limiting reactant will be NaOH.

Now we devise the following reasoning:

if 6 mmoles of NaOH produces 2 mmoles of Al(OH)₃

then 2.2 mmoles of NaOH produces X mmoles of Al(OH)₃

X = (2.2 × 2) / 6 = 0.73 mmoles of Al(OH)₃

mass of Al(OH)₃ = number of moles / molecular weight

mass of Al(OH)₃ = 0.73 / 78

mass of Al(OH)₃ = 9.4 × 10⁻³ mg

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7 0

3 years ago

Which statement is true about Air at different temperatures

Alchen [17]

Are there any choices?

6 0

4 years ago

Read 2 more answers