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2 years ago

What is the outre layer and large part of the sun´s atmosphere

Chemistry

1 answer:

Nadusha1986 [10]2 years ago

7 0

Answer:

Our Sun. Corona – The outer layer of sun's atmosphere. Sunspot – A dark area of gas on the sun's surface that is cooler than surrounding gases. Prominence – A huge, reddish loop of gas that protrudes from the sun's surface, linking part of sunspot regions.

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What is the volume of an object with a mass of 11.5 g and a density of 1.3 g/mL?

Sliva [168]

Answer:

The answer is

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

$volume = \frac{mass}{density} \\$

From the question

mass of object = 11.5 g

density = 1.3 g/mL

The volume is

$volume = \frac{11.5}{1.3} \\ = 8.846153$

We have the final answer as

Hope this helps you

3 0

3 years ago

Given an activity series in which the most active metals are at the top of the list and the least active metals are at the botto

Ugo [173]

Answer:

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Explanation:

6 0

3 years ago

1. Are ions involved in metallic bonding?<br> 2. How are the atoms arranged in a metal? Why?

Katen [24]

Answer:

1 ans =The atoms that the electrons leave behind become positive ions, and the interaction between such ions and valence electrons gives rise to the cohesive or binding force that holds the metallic crystal together.

7 0

2 years ago

Stochiometry .How many grams of CO2 is produced when excess CS2 reacts with 4 mols of O2? CS2+O2-SO2 Please balance equation.

Travka [436]

Mass of CO₂ produced : 58.67 g

<h3>Further explanation</h3>

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

Reaction

CS₂ + 3O₂ -------> CO₂ + 2SO₂

mol of CO₂ based on mol of O₂ as a limiting reactant(CS₂ as an excess reactant)

From the equation, mol ratio of mol CO₂ : mol O₂ = 1 : 3, so mol CO₂ :

$\tt \dfrac{1}{3}\times 4=\dfrac{4}{3}$

mass CO₂ (MW= 44 g/mol) :

$\tt \dfrac{4}{3}\times 44=58.67~g$

4 0

2 years ago

Identify the element oxidized, the element reduced, the oxidizing agent, and the reducing agent for the following equation

bazaltina [42]

The element that gains electrons, becomes reduced.

While the one which loses electrons, becomes oxidized.

In this equation,

CH₃OH + Cr₂O₇²⁻---- --> CH₂O + Cr³⁺.

By balancing the equation, we will get:

3CH₃OH + Cr₂O₇²⁻ + 8H⁺ --> 3CH₂O + 2Cr³⁺ + 7H₂O

Here the oxidation state of Cr changes from +6 to +3 that is it is being reduced thus serving as a oxidizing agent while other element retain their charges.

Here Cr₂O₇²⁻ is reduced while CH₃OH is oxidized.

So Cr₂O₇²⁻ serves as a oxidizing agent, while CH₃OH serves as reducing agent .

4 0

2 years ago