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xxTIMURxx [149]
3 years ago
9

In an experiment, 2.54 grams of copper completely reacts with sulfur, producing 3.18 grams of copper(I) sulfide.

Chemistry
2 answers:
kozerog [31]3 years ago
5 0

Answer:

               0.64 g of S

Solution:

               The balance chemical equation is as follow,

                                           2 Cu + S ----> Cu₂S

According to equation,

                        127 g (2 mole) Cu produces = 159 g (1 mole) of Cu₂S

So,

                                 2.54 g Cu will produce = X g of Cu₂S

Solving for X,

                     X = (2.54 g * 159 g) / 127 g

                     X = 3.18 g of Cu₂S

Now, it is confirmed that the reaction is 100% ideal. Therefore,

As,

                       127 g (2 mole) Cu required = 32 g (1 mole) of S

So,

                                2.54 g Cu will require = X g of S

Solving for X,

                      X = (2.54 g * 32 g) / 127 g

                      X = 0.64 g of S

____ [38]3 years ago
3 0

Answer : The total mass of sulfur consumed will be, 0.64 grams and the chemical formula of the compound produced is, Cu_2S copper oxide.

Solution :

According to the law of conservation of mass, the total mass of reactant should be equal to the total mass of product.

The balanced chemical reaction will be,

2Cu+S\rightarrow Cu_2S

In this reaction, copper and sulfur are the reactants and copper sulfide is the product.

Let the mass of sulfur be 'x' gram

\text{The total mass of reactants}=\text{Total mass of copper}+\text{Total mass of sulfur}=2.54g+x

\text{The total mass of product}=\text{Total mass of copper sulfide}=3.18g

According to the law of conservation of mass,

\text{The total mass of reactants}=\text{The total mass of product}

2.54g+x=3.18g

x=(3.18-2.54)g

x=0.64g

Therefore, the total mass of sulfur consumed will be, 0.64 grams and the chemical formula of the compound produced is, Cu_2S copper oxide.

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Consider the reaction: 2BrF3(g) --> Br2(g) + 3F2(g)
riadik2000 [5.3K]

Answer : The entropy change of reaction for 1.62 moles of BrF_3 reacts at standard condition is 217.68 J/K

Explanation :

The given balanced reaction is,

2BrF_3(g)\rightarrow Br_2(g)+3F_2(g)

The expression used for entropy change of reaction (\Delta S^o) is:

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{Br_2}\times \Delta S_f^0_{(Br_2)}+n_{F_2}\times \Delta S_f^0_{(F_2)}]-[n_{BrF_3}\times \Delta S_f^0_{(BrF_3)}]

where,

\Delta S^o = entropy change of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

\Delta S_f^0_{(Br_2)} = 245.463 J/mol.K

\Delta S_f^0_{(F_2)} = 202.78 J/mol.K

\Delta S_f^0_{(BrF_3)} = 292.53 J/mol.K

Now put all the given values in this expression, we get:

\Delta S^o=[1mole\times (245.463J/K.mole)+3mole\times (202.78J/K.mole)}]-[2mole\times (292.53J/K.mole)]

\Delta S^o=268.74J/K

Now we have to calculate the entropy change of reaction for 1.62 moles of BrF_3 reacts at standard condition.

From the reaction we conclude that,

As, 2 moles of BrF_3 has entropy change = 268.74 J/K

So, 1.62 moles of BrF_3 has entropy change = \frac{1.62}{2}\times 268.74=217.68J/K

Therefore, the entropy change of reaction for 1.62 moles of BrF_3 reacts at standard condition is 217.68 J/K

3 0
4 years ago
Based on nuclear stability, what is the symbol for the most likely product nuclide when iodine-123 undergoes decay?
MissTica
Answer is: Te (tellurium).
During electron capture, iodine-123 f<span>orm the nearly-stable nuclide tellurium-123.
</span>In beta plus decay (atomic number Z is decreased by one), a proton is converted to a neutron and positron and an electron neutrino, so mass <span>number does not change.
</span>Iodine-123 is a radioactive isotope of iodine used in nuclear medicine imaging.<span>
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6 0
3 years ago
Which of these molecular geometries forms a molecule with a bent shape?
allsm [11]

Answer:

well plus da cuh cause opp homie freak jit.on 390/it = 30

3 0
2 years ago
When the following oxidation-reduction occurs, what is the balanced reduction half-reaction after the electrons in both half rea
Lostsunrise [7]

Answer :  The balanced reduction half-reaction is:

3Cu^{2+}+6e^-\rightarrow 3Cu

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

The given balanced redox reaction is :

Al(s)+Cu^{2+}(aq)\rightarrow Al^{3+}(aq)+Cu(s)

The half oxidation-reduction reactions are:

Oxidation reaction : Al\rightarrow Al^{3+}+3e^-

Reduction reaction : Cu^{2+}+2e^-\rightarrow Cu

In order to balance the electrons, we multiply the oxidation reaction by 2 and reduction reaction by 3 and then added both equation, we get the balanced redox reaction.

Oxidation reaction : 2Al\rightarrow 2Al^{3+}+6e^-

Reduction reaction : 3Cu^{2+}+6e^-\rightarrow 3Cu

The balanced redox reaction will be:

2Al(s)+3Cu^{2+}(aq)\rightarrow 2Al^{3+}(aq)+3Cu(s)

Thus, the balanced reduction half-reaction is:

3Cu^{2+}+6e^-\rightarrow 3Cu

6 0
3 years ago
How many oxygen atoms are present in 4.28 g of LiBroz sample?
Dafna11 [192]

Answer:

0.43×10²³ atoms

Explanation:

Given data:

Mass of LiBrO₂ = 4.28 g

Number of atoms of oxygen = ?

Solution:

Number of moles = mass/molar mass

Number of moles = 4.28 g/ 118.84 g/mol

Number of moles = 0.036 mol

We can see 1 mole of LiBrO₂ contain 2 mole of oxygen atm.

0.036 mol × 2 = 0.072 mol

1 mole contain 6.022×10²³ atoms

0.072 mol × 6.022×10²³ atoms / 1mol

0.43×10²³ atoms

3 0
3 years ago
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