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2 years ago

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what is the scientific name for the layer of gas that surrounds the earth A. hydrosphere B. biosphere C. atmosphere D. lithosphe

re

2 answers:

horsena [70]2 years ago

7 0

Answer:

hydro

Explanation: hydrogen

Marat540 [252]2 years ago

7 0

Answer:

C. Atmosphere

Explanation:

First, let’s define each answer choice.

Hydrosphere: all of the water on Earth

Biosphere: the global ecosystem where organisms live

Atmosphere: the envelope of gases around the planet

Lithosphere: the outermost part of the Earth( the crust and upper mantle)

The question asks us to find the scientific name for the layer of gas that surrounds the Earth. The best choice is C. Atmosphere.

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Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

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

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

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

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

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

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

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

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

2 years ago

What is the molar mass of KOH?

Julli [10]

<h3>Answer:</h3>

56.11 g/mol

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Compound] KOH

<u>Step 2: Identify</u>

[PT] Molar Mass of K - 39.10 g/mol

[PT] Molar Mass of O - 16.00 g/mol

[PT] Molar Mass of H - 1.01 g/mol

<u>Step 3: Find</u>

39.10 + 16.00 + 1.01 = 56.11 g/mol

5 0

2 years ago

Antarctica, almost completely covered in ice, has an area

Marysya12 [62]

<u>Answer:</u> The mass of ice is $2.39\times 10^{22}g$

<u>Explanation:</u>

We are given:

Area of Antarctica = $5,500,000mi^2=5,500,000\times 2.59\times 10^{10}=142.45\times 10^{15}cm^2$ (Conversion factor: $1mi^2=2.59\times 10^{10}cm^2$ )

Height of Antarctica with ice = 7500 ft.

Height of Antarctica without ice = 1500 ft.

Height of ice = 7500 - 1500 = 6000 ft = $182.88\times 10^3cm$ (Conversion factor: 1 ft = 30.48 cm)

To calculate mass of ice, we use the equation:

$\text{Density of a substance}=\frac{\text{Mass of a substance}}{\text{Volume of a substance}}$

We are given:

Density of ice = $0.917g/cm^3$

Volume of ice = Area × Height of ice = $142.45\times 10^{15}cm^2\times 182.88\times 10^3cm=26051.26\times 10^{18}cm^3$

Putting values in above equation, we get:

$0.917g/cm^3=\frac{\text{Mass of ice}}{26051.26\times 10^{18}cm^3}\\\\\text{Mass of ice}=(0.917g/cm^3\times 26051.26\times 10^{18}cm^3=2.39\times 10^{22}g$

Hence, the mass of ice is $2.39\times 10^{22}g$

5 0

2 years ago

The law of conservation of mass is applicable to

katrin2010 [14]

Answer:

The law of conservation of mass states that in a closed system, mass is neither created nor destroyed during a chemical or physical reaction. The law of conservation of mass is applied whenever you balance a chemical equation.

Explanation:

According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.

The law of conservation of mass is useful for a number of calculations and can be used to solve for unknown masses, such the amount of gas consumed or produced during a reaction.

It is applicable in a chemical when the the mass of the products in a chemical reaction is equal to the mass of the reactants.

But it is not applicable in a nuclear fusion as some of the mass is generated as energy.

3 0

2 years ago

You are working the grill at a restaurant and are having trouble getting the grill lit through electrical means. You light a mat

never [62]

Answer:

in this situation I would a little bold

Explanation:

first I don't know what extinguisher I would use pretty much any that helps with fires. I'll back people up, take the hood and put it on the small fire that way it will light out more and if I open the hood and there still a little fire I would use the extinguisher and no one gets hurt :)

6 0

2 years ago