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katovenus [111]

3 years ago

13

Layer B in the diagram is located between the crust and core in terms of location, temperature, and pressure. What part of the e

arth is labeled layer B?
A) the core
B) the crust
C) the mantle
D) the atmosphere

2 answers:

nignag [31]3 years ago

8 0

The answer is c) the mantle

Slav-nsk [51]3 years ago

3 0

if u in studyisland the answer is A. core

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What can you summarize about the use of acid-base indicators?

densk [106]

An acid-base indicator is used to identify the presence of an acid or base. These indicators exhibit different colors depending on the solution that they are in. They are especially useful when completing titrations to determine the molarity of an unknown substance and is denoted as option D.

<h3>What is Titration?</h3>

This is done in the laboratory and involves the slow addition of one solution of a known concentration to a known volume of another solution of unknown concentration.

Acid-base indicators are used to determine the presence of an acid or base in a solution which is based on the colors seen when performing the chemical reaction.

It is used to calculate the the molarity of an unknown substance through the knowledge of the other parameters which is therefore the reason why option D was chosen as the most appropriate choice.

Read more Acid-base indicators here brainly.com/question/2815636

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6 0

10 months ago

A mixture of ethyl acetate vapour and air has a relative saturation of 50% at 303 K and a total pressure of 100 kPa. If the vapo

AURORKA [14]

Answer : The correct option is, (b) 0.087

Explanation :

The formula used for relative saturation is:

$\text{Relative saturation}=\frac{P_A}{P_A^o}$

where,

$P_A$ = partial pressure of ethyl acetate

$P_A^o$ = vapor pressure of ethyl acetate

Given:

Relative saturation = 50 % = 0.5

Vapor pressure of ethyl acetate = 16 kPa

Now put all the given values in the above formula, we get:

$0.5=\frac{P_A}{16kPa}$

$P_A=8kPa$

Now we have to calculate the molar saturation.

The formula used for molar saturation is:

$\text{Molar saturation}=\frac{P_{vapor}}{P_{\text{vapor free}}}$

and,

P(vapor free) = Total pressure - Vapor pressure

P(vapor) = $P_A$ = 8 kPa

So,

P(vapor free) = 100 kPa - 8 kPa = 92 kPa

The molar saturation will be:

$\text{Molar saturation}=\frac{P_{vapor}}{P_{\text{vapor free}}}$

$\text{Molar saturation}=\frac{8kPa}{92kPa}=0.087$

Therefore, the molar saturation is 0.087

5 0

3 years ago

Branches of chemistry with definition and examples in real life

nataly862011 [7]

Explanation:The five major branches of chemistry are organic, inorganic, analytical, physical, and biochemistry.

...

Sub-branches of physical chemistry include:

Photochemistry — the study of the chemical changes caused by light.

Surface chemistry — the study of chemical reactions at surfaces of substances

8 0

2 years ago

What happened to the concentration of the product of H2O when the reaction shifts to the left

lilavasa [31]

I might need a diagram for this, but I have a vague idea of what you are talking about.

If H20 is going left it means the temperature is going lower.

The molecules will condense to slowly become ice

6 0

3 years ago

Hydrogen and Methanol have both been proposed as alternatives to hydrocarbon fuels. Write balanced reactions for the complete co

Arlecino [84]

Answer:

The order of energy released per mass is

CH₃OH (-2.268 × 10⁴ kJ/kg) < C₈H₁₈ (-4.826 × 10⁴ kJ/kg) < H₂ (-2.835 × 10⁵ kJ/kg)

Explanation:

In order to calculate the enthalpy of a reaction (ΔH°r) we can use the following expression.

ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)

where

ΔH°f(i) are the enthalpies of formation of reactants and products

ni are the moles of reactants and products

<u>Combustion of hydrogen</u>

H₂(g) + 1/2 O₂(g) ⇒ H₂O(l)

ΔH°r = 2 mol × ΔH°f(H₂O) - 1 mol × ΔH°f(H₂) - 1/2 mol × ΔH°f(O₂)

ΔH°r = 2 mol × (-285.8 kJ/mol) - 1 mol × 0 - 1/2 mol × 0

ΔH°r = -571.6 kJ

571.6 kJ are released when 1 mole of H₂ is burned. The amount of heat released per kilogram is:

$\frac{-571.6kJ}{1molH_{2}} .\frac{1molH_{2}}{2.016gH_{2}} .\frac{10^{3}gH_{2} }{1kgH_{2}} =-2.835 \times 10^{5} kJ/kgH_{2}$

<u>Combustion of methanol</u>

CH₃OH(l) + 3/2 O₂(g) ⇒ CO₂(g) + 2 H₂O(l)

ΔH°r = 1 mol × ΔH°f(CO₂) + 2 mol × ΔH°f(H₂O) - 1 mol × ΔH°f(CH₃OH) - 3/2 mol × ΔH°f(O₂)

ΔH°r = 1 mol × (-393.5 kJ/mol) + 2 mol × (-285.8 kJ/mol) - 1 mol × (-238.4 kJ/mol) - 3/2 mol × 0

ΔH°r = -726.7 kJ

726.7 kJ are released when 1 mole of CH₃OH is burned. The amount of heat released per kilogram is:

$\frac{-726.7kJ}{1molCH_{3}OH} .\frac{1molCH_{3}OH}{32.04gCH_{3}OH} .\frac{10^{3}gCH_{3}OH }{1kgCH_{3}OH} =-2.268 \times 10^{4} kJ/kgCH_{3}OH$

<u>Combustion of octane</u>

C₈H₁₈(l) + 12.5 O₂(g) ⇒ 8 CO₂(g) + 9 H₂O(l)

ΔH°r = 8 mol × ΔH°f(CO₂) + 9 mol × ΔH°f(H₂O) - 1 mol × ΔH°f(C₈H₁₈) - 12.5 mol × ΔH°f(O₂)

ΔH°r = 8 mol × (-393.5 kJ/mol) + 9 mol × (-285.8 kJ/mol) - 1 mol × (-208.4 kJ/mol) - 12.5 mol × 0

ΔH°r = -5511.8 kJ

5511.8 kJ are released when 1 mole of C₈H₁₈ is burned. The amount of heat released per kilogram is:

$\frac{-5511.8kJ}{1molC_{8}H_{18}} .\frac{1molC_{8}H_{18}}{114.2gC_{8}H_{18}} .\frac{10^{3}gC_{8}H_{18} }{1kgC_{8}H_{18}} =-4.826 \times 10^{4} kJ/kgC_{8}H_{18}$

5 0

3 years ago