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3 years ago

What are inside of cars so hot when left in the sun

Chemistry

2 answers:

Naya [18.7K]3 years ago

8 0

Answer:

metalllll/material

Explanation:

you know how dark shirts absorb the sun heat well its practically the same just with material.

"Why is the inside of a car so hot after sitting in the sun?

Cars warm up in the sun due to the greenhouse effect: Sunlight passing through the windows into the car is mostly absorbed by interior surfaces, then radiated back to the air as heat. ... “As a consequence, the inside of the car will warm because radiation is coming in but not much is going back out.”Jul 27, 2009

Curiosities: Does a dark-colored car heat up more in the sun than a ..."

Alchen [17]3 years ago

8 0

^^ Also is caused by something called the Greenhouse effect which causes ultra violet rays to come in and be reflected with infrared rays which get trapped and therefore cause heat for the car to be hot since infrared radiation is another word for heat

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Which two functional groups are always found in amino acids?

Dmitry_Shevchenko [17]

The answer would be the amino group and the carboxyl group
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6 0

3 years ago

Yeast converts glucose to ethanol and carbon dioxide during a process known as anaerobic fermentation. The chemical reaction is

Korolek [52]

Answer:

102g

Explanation:

To find the mass of ethanol formed, we first need to ensure that we have a balanced chemical equation. A balanced chemical equation is where the number of atoms of each element is the same on both sides of the equation (reactants and products). This is useful as only when a chemical equation is balanced, we can understand the relationship of the amount (moles) of reactant and products, or to put it simply, their relationship with one another.

In this case, the given equation is already balanced.

$\tex{C_6H_{12}O_6} \longrightarrow 2 \ C_2H_6O + 2 \ CO_2$

From the equation, the amount of ethanol produced is twice the amount of yeast present, or the same amount of carbon dioxide produced. Do note that amount refers to the number of moles here.

Mole= Mass ÷Mr

Mass= Mole ×Mr

Method 1: using the mass of glucose

Mr of glucose

= 6(12) +12(1) +6(16)

= 180

Moles of glucose reacted

= 200 ÷180

= $\frac{10}{9}$ mol

Amount of ethanol formed: moles of glucose reacted= 2: 1

Amount of ethanol

= $2(\frac{10}{9} )$

= $\frac{20}{9}$ mol

Mass of ethanol

= $\frac{20}{9} \times[2(12)+6+16]$

= $\frac{20}{9}(46)$

= 102 g (3 s.f.)

Method 2: using mass of carbon dioxide produced

Mole of carbon dioxide produced

= 97.7 ÷[12 +2(16)]

= 97.7 ÷44

= $\frac{977}{440}$ mol

Moles of ethanol: moles of carbon dioxide= 1: 1

Moles of ethanol formed= $\frac{977}{440}$ mol

Mass of ethanol formed

= $\frac{977}{440} \times[2(12)+6+16]$

= 102 g (3 s.f.)

Thus, 102 g of ethanol are formed.

Additional:

For a similar question on mass and mole ratio, do check out the following!

brainly.com/question/1685725

8 0

1 year ago

What element was named for the scientist who discovered the nucleus of the atom using gold foil

Artist 52 [7]

rutherfordium element # 104

4 0

3 years ago

Express each of the following in standard form.

swat32

Answer:3.6 x 101 or 8.77 x 10-1

3 0

3 years ago

Consider the following system at equilibrium:A(aq)+B(aq) <---> 2C(aq)Classify each of the following actions by whether it

velikii [3]

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On addition of reactant at equilibrium shifts the equilibrium in forward direction.
On addition of product at equilibrium shifts the equilibrium in backward direction.
On removal of reactant at equilibrium shifts the equilibrium in backward direction.
On removal of product at equilibrium shifts the equilibrium in forward direction.

$A(aq)+B(aq)\rightleftharpoons 2C(aq)$

Reactants = A , B

Product = C

1. Increase A

On increasing the amount of A at equilibrium will shift the equilibrium in forward or rightward direction.

2. Increase B

On increasing the amount of B at equilibrium will shift the equilibrium in forward or rightward direction.

3. Increase C

On increasing the amount of C at equilibrium will shift the equilibrium in backward or leftward direction.

4. Decease A

On decreasing the amount of A at equilibrium will shift the equilibrium in backward or leftward direction.

5. Decease B

On decreasing the amount of B at equilibrium will shift the equilibrium in backward or leftward direction.

6. Decease C

On decreasing the amount of C at equilibrium will shift the equilibrium in forward or rightward direction.

7. Double A and Halve B

Equilibrium constant of the reaction = K

$K=\frac{[C]^2}{[A][B]}$

On doubling A and halving B, equilibrium constant of the reaction = K'

$K'=\frac{[C]^2}{[2A][\frac{B}{2}]}=\frac{[C]^2}{[A][B]}$

The value of equilibrium constant K' is equal to K, which means that equilibrium will not shift in any direction.

8. Double both B and C

Equilibrium constant of the reaction = K

$K=\frac{[C]^2}{[A][B]}$

On doubling B and C, equilibrium constant of the reaction = K'

$K'=\frac{[2C]^2}{[A][2B]}=\frac{4[C]^2}{[A][2B]}=\frac{2[C]^2}{[A][B]}$

K' = 2 K

The value of equilibrium constant K' is double the K, which means that product is increasing which means that equilibrium will shift in backward or leftward direction.

5 0

3 years ago