Write the complete balanced equation for Manganese & sulfuric acid --> Manganese (II) sulfate & water & sulfur dioxide no spaces
no subscripts
no 1's for coefficients

WILL GIVE BRAINLIEST!

Answers

Answer 1

Answer:

Answer:

I kinda forgot. I'm sorry if I didn't answer your question.

Explanation:

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What are the trends for ionic size among the cations and anions?

Answers

Cations are smaller than the atoms from which they are formed. Anions are larger than the atoms from which they were formed. Ionic size increases from top to bottom down a group of elements in the periodic table.

If an element, X, can form an oxide that has theformula X2O3, then element X would most likely be located on the Periodic Table in the same group as(1) Ba (3) In
(2) Cd (4) Na

Answers

Answer: (3) In

Explanation: As the given compound is $X_2O_3$ , it can be said that X has a valency of 3.

1)Barium (Ba): Atomic no: 56: $Xe6s^2$ : In order to complete its octet, it needs to donate 2 electrons, and thus the valency is 2.

2) Cadmium (Cd) : Atomic no: 48: $Kr5s^24d^(10)$ : In order to complete its octet, it needs to donate 2 electrons, and thus the valency is 2.

3) Indium (In): Atomic no: 49: $Kr5s^25p^(1)$ : In order to complete its octet, it needs to donate 3 electrons, and thus the valency is 3.

4) Sodium (Na): Atomic no: 11: $Ne3s^1$ : In order to complete its octet, it needs to donate 1 electron, and thus the valency is 1.

So the given element could only belong to the group containing Indium (In) as all the elements in the same group exhibit similar valency.

In! This is becasue it has a valency of 3

If 200.0 g of copper(II) sulfate react with an excess of zinc metal, what is the theoretical yield of copper?

Answers

Answer:

The theoretical yield of copper is 79.6281 g

Explanation:

$CuSO_4+Zn\rightarrow ZnSO_4+Cu$

Moles of copper sulfate = $(200.0 g)/(159.609 g/mol)=1.2530 mol$

According to reaction 1 mol of copper sulfate gives 1 mol of copper .

Then 1.2530 mol of copper sulfate will gives :

$(1)/(1)* 1.2530 mol=1.2530 mol$ of copper .

Mass of 1.2530 mol of copper :

1.2530 mol × 63.55 g/mol = 79.6281 g

The theoretical yield of copper is 79.6281 g

The theoretical yield of copper is a 78:32 ratio.

Which factor is generally responsible for high melting points?

Answers

Strong intermolecular ion-ion forces are responsible for high melting points. Thus the solid state requires close proximity and high order to have high melting point. Hope this answer helps. Good luck!

Answer;

Intermolecular forces.

Explanation;

Melting point is the temperature at which a solid melts or a solid changes to a liquid. In liquids melting point is similar to the freezing point.
Melting point of a substance depends on the molecular composition, force of attraction between molecules and also the presence of impurities.
A strong force of attraction between molecules results in a higher melting point. Such that stronger intermolecular forces results to a higher melting point.
For example, the melting point of ionic compounds is higher due to the strong electrostatic forces joining ions.

Which of the following is true of solids?a. All solids have equal melting points.
b. Ionic solids have higher melting points than molecular solids.
c. Molecular solids have higher melting points than all other types of solids.
d. It is impossible for solids to melt; therefore solids do not have melting points.

Answers

B. Ionic solids have higher melting points than molecular solids.

A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. Which is thelimiting reactant and how much excess reactant remains after the reaction
has stopped?

Answers

Answer:

Ammonia is limiting reactant

Amount of oxygen left = 0.035 mol

Explanation:

Given data:

Masa of ammonia = 2.00 g

Mass of oxygen = 4.00 g

Which is limiting reactant = ?

Excess reactant's amount left = ?

Solution:

Balance chemical equation:

4NH₃ + 3O₂ → 2N₂ + 6H₂O

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 2.00 g/ 17 g/mol

Number of moles = 0.12 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 4.00 g/ 32 g/mol

Number of moles = 0.125 mol

Now we will compare the moles of ammonia and oxygen with water and nitrogen.

NH₃ : N₂

4 : 2

0.12 : 2/4×0.12 = 0.06

NH₃ : H₂O

4 : 6

0.12 : 6/4×0.12 = 0.18

O₂ : N₂

3 : 2

0.125 : 2/3×0.125 = 0.08

O₂ : H₂O

3 : 6

0.125 : 6/3×0.125 = 0.25

The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.

Amount of oxygen left:

NH₃ : O₂

4 : 3

0.12 : 3/4×0.12= 0.09

Amount of oxygen react = 0.09 mol

Amount of oxygen left = 0.125 - 0.09 = 0.035 mol

Final answer:

The limiting reactant in the reaction between Ammonia and Oxygen is Ammonia (NH3). All of the Oxygen is used up in the reaction, so no excess reactant remains.

Explanation:

This question involves a concept in chemistry known as limiting reactants and stoichiometry. The balanced chemical reaction between Ammonia (NH3) and Oxygen (O2) is: 4NH3 + 5O2 -> 4NO + 6H2O. This indicates that 4 moles of NH3 react with 5 moles of O2.

To find the limiting reactant, you first need to convert the grams of your reactants to moles. The molar mass of NH3 is approximately 17.0g/mol, and the molar mass of O2 is 32.0g/mol. Therefore, you have 2.00g/17.0g/mol = 0.118 moles of NH3 and 4.00g/32.0g/mol = 0.125 moles of O2.

Since 5 moles of O2 are needed for every 4 moles of NH3, and we have slightly more O2 than NH3, the limiting reactant is NH3. To find the amount of excess reactant, we determine how much O2 actually reacted by multiplying (0.118 moles NH3)*(5 moles O2/4 moles NH3) = 0.1475 moles O2. The original amount of O2 was 0.125 moles, so the amount left over is 0.125 - 0.1475, which is a negative number and thus not possible. This confirms that O2 is the excess reactant, although it entirely reacted. Hence, no excess reactant remains.