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Anastasy [175]
3 years ago
14

NEED HELP ASAP PLEASE!!! EMERGENCY!

Chemistry
2 answers:
N76 [4]3 years ago
4 0
D I believe peace out Hope this helps
PolarNik [594]3 years ago
3 0
I think its C but im not sure

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Identify the oxidation state of Ba 2 + . +2 Identify the oxidation state of S in SO 2 . +2 Identify the oxidation state of S in
cupoosta [38]

Answer:

The oxidation state of Ba in cation Ba²⁺ is +2

The oxidation state of S in SO₂, is +4

The oxidation state of S in anion sulfate (SO₄⁻²) is +6

The oxidation state of Zn in the Zinc sulfate, is +2

Explanation:

We define oxidation state as the number which can be negative or positive that  

indicates the number of electrons that the atom has accepted or transferred.

All the elements in ground state has 0 as oxidation state.

This numbers are very important for redox reaction which are balanced by the ion electron method.

When the elements gain electrons, the element is being reduced so the oxidation state decreases.

When the elements release electrons, the element is oxidized so the oxidation state increases.

We have to think, that global charge of a compound is 0, for example in the ZnSO₄.

The sulfate anion has a global charge of -2 because it has released 2 protons, it came from the sulfuric acid (H₂SO₄). As the global charge is -2, oxygen acts with -2, and the anion has 4 atoms so the global charge of O is -8. Definetly S, has +6 as oxidation state.

In the SO₂, oxygen acts with -2 and there are 2 atoms in the compound, so the global charge is 0 and the global charge for O  is -4. Therefore S must act with +4.

Ba²⁺ is an element of group 2 and has a tendency to form a cation, so it can release electrons for that purpose.  At least, it can release 2 e⁻, that's why the oxidation state is +2. It can complete the octet rule and it will be isoelectronic with Xe.

3 0
3 years ago
Conservation of matter article questions
elena-14-01-66 [18.8K]

The Law of conservation of mass states that  option C: matter is neither created nor destroyed.

<h3>What is the law of conservation of matter?</h3>

Physical and chemical changes can cause matter to transform into different forms, but no matter what happens, matter is always conserved. There is no creation or destruction of matter; the amount of matter is the same before and after the transformation.

The principle of matter conservation. argues that matter cannot be generated or destroyed during a chemical reaction. The same number of atoms exist before and after the alterations even though the matter may shift from one form to another. reactant.

Therefore, According to the principle of mass conservation, neither chemical processes nor physical changes can create or destroy mass in an isolated system. The mass of the products and reactants of a chemical reaction must be equal, in accordance with the law of conservation of mass.

Learn more about matter from

brainly.com/question/3998772
#SPJ1

See full question below

1. Multiple-choice

Q.

Conservation of matter article questions

Law of conservation of mass states that

answer choices

matter is created

matter is destroyed

matter is neither created nor destroyed

matter does not change

3 0
1 year ago
Explain the change in mass that occurs when the following substances are separately heated in open crucibles
aleksandrvk [35]

Answer:

your answer is (a) Copper Metal

Explanation:

7 0
3 years ago
Which of the following is not the same as 12.5 millimeters
jek_recluse [69]
I got <span>0.00125 hm . Hope I helped, and good luck(: </span>
5 0
3 years ago
Assume the hydrolysis of ATP proceeds with ΔG′° = –30 kJ/mol. ATP + H2O → ADP + Pi Which expression gives the ratio of ADP to AT
Andru [333]

Answer:

6.14\cdot 10^{-6}

Explanation:

Firstly, write the expression for the equilibrium constant of this reaction:

K_{eq} = \frac{[ADP][Pi]}{ATP}

Secondly, we may relate the change in Gibbs free energy to the equilibrium constant using the equation below:

\Delta G^o = -RT ln K_{eq}

From here, rearrange the equation to solve for K:

K_{eq} = e^{-\frac{\Delta G^o}{RT}}

Now we know from the initial equation that:

K_{eq} = \frac{[ADP][Pi]}{ATP}

Let's express the ratio of ADP to ATP:

\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}}

Substitute the expression for K:

\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}}

Now we may use the values given to solve:

\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}} = [Pi]e^{\frac{\Delta G^o}{RT}} = 1.0 M\cdot e^{\frac{-30 kJ/mol}{2.5 kJ/mol}} = 6.14\cdot 10^{-6}

7 0
3 years ago
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