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Mazyrski [523]
2 years ago
15

All of the following reactions can be described as displacement reactions except:____________.

Chemistry
1 answer:
Lady_Fox [76]2 years ago
8 0

Answer:

b

Explanation:

The reaction that is not a displacement reaction from all the options is C_6H_6_{(l)} + Cl_{2(g)} --> C_6H_5Cl_{(l)} + HCl_{(g)}

In a displacement reaction, a part of one of the reactants is replaced by another reactant. In single displacement reactions, one of the reactants completely displaces and replaces part of another reactant. In double displacement reaction, cations and anions in the reactants switch partners to form products.

<em>Options a, c, d, and e involves the displacement of a part of one of the reactants by another reactant while option b does not.</em>

Correct option = b.

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The are eight hydrogen atoms in ammonium sulfide because there are 2 molecules of ammonium.
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An archeologist finds a 1.62 kg goblet that she believes to be made of pure gold. She determines that the volume of the goblet i
katovenus [111]
The  answer is

<span>The density (D) is quotient of mass (m) and volume (V):
</span>D= \frac{m}{V}
The unit is g/cm³

It is given:
m = 1.62 kg = 1620 g
V = 205 mL = 205 cm³
D = ?

Thus:
D= \frac{m}{V} = \frac{1620g}{205cm ^{3} } = 7.90 g/cm ^{3}

The density of the goblet is 7.90 g/cm³.
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2 years ago
At home, Helena is cooking with butter and lard. While the fats sit on the counter, she notices that the butter begins to melt,
pychu [463]

Answer: Lard contains the most hydrogen atoms

Explanation:

Unlike butter, lard contains more saturated fats, higher monounsaturated fats, and no trans fat (a type of unsaturated fat).

Thus, the presence of more saturated fats allows for more hydrogen atoms to confer heat stability to lard.

So, lard contains more hydrogen atoms

4 0
2 years ago
A solution is made by dissolving 58.125 g of sample of an unknown, nonelectrolyte compound in water. The mass of the solution is
e-lub [12.9K]

Answer:

molecular weight (Mb) = 0.42 g/mol

Explanation:

mass sample (solute) (wb) = 58.125 g

mass sln = 750.0 g = mass solute + mass solvent

∴ solute (b) unknown nonelectrolyte compound

∴ solvent (a): water

⇒ mb = mol solute/Kg solvent (nb/wa)

boiling point:

  • ΔT = K*mb = 100.220°C ≅ 373.22 K

∴ K water = 1.86 K.Kg/mol

⇒ Mb = ? (molecular weight) (wb/nb)

⇒ mb = ΔT / K

⇒ mb = (373.22 K) / (1.86 K.Kg/mol)

⇒ mb = 200.656 mol/Kg

∴ mass solvent = 750.0 g - 58.125 g = 691.875 g = 0.692 Kg

moles solute:

⇒ nb = (200.656 mol/Kg)*(0.692 Kg) = 138.83 mol solute

molecular weight:

⇒ Mb = (58.125 g)/(138.83 mol) = 0.42 g/mol

8 0
2 years ago
Coal gasification is a multistep process to convert coal into cleaner-burning fuels. In one step, a coal sample reacts with supe
ddd [48]

Answer :

The enthalpy of reaction is, -187.6 kJ/mol

The total heat will be, -2251 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

(a) The formation of CH_4 will be,

2C(coal)+2H_2O(g)\rightarrow CH_4(g)+CO_2(g)    \Delta H_{rxn}=?

The intermediate balanced chemical reaction will be,

(1) C(coal)+H_2O(g)\rightarrow CO(g)+H_2(g)     \Delta H_1=29.7kJ

(2) CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)    \Delta H_2=-41kJ

(3) CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)    \Delta H_3=-206kJ

We are multiplying equation 1 by 2 and then adding all the equations, we get :

(b) The expression for enthalpy of reaction will be,

\Delta H_{rxn}=2\times \Delta H_1+\Delta H_2+\Delta H_3

\Delta H_{rxn}=(2\times 29.7)+(-41)+(-206)

\Delta H_{rxn}=-187.6kJ/mol

Therefore, the enthalpy of reaction is, -187.6 kJ/mol

(c) Now we have to calculate the total heat.

\Delta H=\frac{q}{n}

or,

q=\Delta H\times n

where,

\Delta H = enthalpy change = -187.6 kJ/mol

q = heat = ?

n = number of moles of coal = \frac{1.00\times 1000g}{12.00g/mol}=83.33mol

Now put all the given values in the above formula, we get:

q=(-187.6kJ/mol)\times (83.33mol)=-2.251kJ

Thus, the total heat will be, -2251 kJ

4 0
2 years ago
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