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3 years ago

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What is the enthalpy in KJ of reaction for the initial combustion if the delta H for vaporization is 75.75 KJ and the net delta

H is -700 kJ?
Write your answer to hundredths, including a negative sign if needed. Do not write the unit.

1 answer:

Gnesinka [82]3 years ago

4 0

Answer:

-775.75

Explanation:

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What is the net ionic equation for the precipitation reaction between BaCl2 and Na2So4? I know the answer ( Ba2+(aq) + SO42-(aq)

elixir [45]

Answer:

Net ionic equation:

Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)

Explanation:

Chemical equation:

BaCl₂ + Na₂SO₄ → BaSO₄ + NaCl

Balanced Chemical equation:

BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)

Ionic equation:

Ba²⁺(aq) + 2Cl⁻(aq) + 2Na⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)+ 2Na⁺(aq) + 2Cl⁻ (aq)

Net ionic equation:

Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)

The Cl⁻(aq) and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The BaSO₄ can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

5 0

3 years ago

A bottle of wine contains 9.81 grams of C2H5OH, dissolved in 87.5 grams of water. The final volume of the solution is 100.0 mL.

dimulka [17.4K]

Answer:

[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)

Explanation:

1. Molarity = moles solute / Volume solution in Liters

=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH

=> volume of solution (assuming density of final solution is 1.0g/ml) ...

volume solution = 9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution

Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH

2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)

From part 1 => [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln

= {(2.2mol)(46g/mol)]/1000g soln] x 100% = 10.1% (w/w) in EtOH.

3 0

3 years ago

A 51.24-g sample of Ba(OH)2 is dissolved in enough water to make 1.20 liters of solution. How many mL of this solution must be d

g100num [7]

Answer:

0.40 L

Explanation:

Calculation of the moles of $Ba(OH)_2$ as:-

Mass = 51.24 g

Molar mass of $Ba(OH)_2$ = 171.34 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{51.24\ g}{171.34\ g/mol}$

$Moles= 0.2991\ mol$

Volume = 1.20 L

The expression for the molarity is:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.2991\ mol}{1.20\ L}=0.24925\ M$

Thus,

Considering

$Molarity_{working\ solution}\times Volume_{working\ solution}=Molarity_{stock\ solution}\times Volume_{stock\ solution}$

Given that:

$Molarity_{working\ solution}=0.100\ M$

$Volume_{working\ solution}=1\ L$

$Volume_{stock\ solution}=?$

$Molarity_{stock\ solution}=0.24925\ M$

So,

$0.100\ M\times 1\ L=0.24925\ M\times Volume_{stock\ solution}$

$Volume_{stock\ solution}=\frac{0.100\times 1}{0.24925}\ L=0.40\ L$

<u>The volume of 0.24925M stock solution added = 0.40 L </u>

6 0

3 years ago

A reaction produces 0.883 moles of H₂O. How many molecules are produced?

prisoha [69]

The answer is 5.32 × 10²³ molecules

<span>Avogadro's number is the number of units (atoms, molecules) in 1 mole of substance:
</span>6.023 <span>× 10²³ units per 1 mole

We have 0.883 moles.
If 1 mole has </span>6.023 × 10²³ molecules, 0.883 moles will have x molecules:
1 mole : 6.023 × 10²³ molecules = 0.883 moles : x

x = 6.023 × 10²³ molecules * 0.883 moles : 1 mole = 5.32 × 10²³ molecules

6 0

3 years ago

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.

sp2606 [1]

Answer:

Mass of CaCl₂ = 20 g

CaCO is presewnt in excess.

Mass of of CaCO₃ remain unreacted = 7.007 g

Explanation:

Given data:

Mass of calcium carbonate = 25 g

Mass of hydrochloric acid = 13.0 g

Mass of calcium chloride produced = ?

Chemical equation:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Number of moles of CaCO₃:

Number of moles of CaCO₃ = Mass /molar mass

Number of moles of CaCO₃= 25.0 g / 100.1 g/mol

Number of moles of CaCO₃ = 0.25 mol

Number of moles of HCl:

Number of moles of HCl = Mass /molar mass

Number of moles of HCl = 13.0 g / 36.5 g/mol

Number of moles of HCl = 0.36 mol

Now we will compare the moles of CaCl₂ with HCl and CaCO₃ .

CaCO₃ : CaCl₂

1 : 1

0.25 : 0.25

HCl : CaCl₂

2 : 1

0.36 : 1/2 × 0.36 = 0.18 mol

The number of moles of CaCl₂ produced by HCl are less it will be limiting reactant.

Mass of CaCl₂ = moles × molar mass

Mass of CaCl₂ =0.18 mol × 110.98 g/mol

Mass of CaCl₂ = 20 g

The calcium carbonate is present in excess.

HCl : CaCO₃

2 : 1

0.36 : 1/2 × 0.36 = 0.18 mol

So, 0.18 moles react with 0.36 moles of HCl.

The moles of CaCO₃ remain unreacted = 0.25 -0.18

The moles of CaCO₃ remain unreacted = 0.07 mol

Mass of of CaCO₃ remain unreacted = Moles × molar mass

Mass of of CaCO₃ remain unreacted = 0.07 mol × 100.1 g/mol

Mass of of CaCO₃ remain unreacted = 7.007 g

7 0

3 years ago