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3 years ago

Label the respiratory system. I’ll mark Brainliest! if y’all help with all of them!!

Chemistry

1 answer:

Ann [662]3 years ago

7 0

Its kinda hard to see what the numbers are pointing to but

1:nasal cavity
2:mouth
3:larynx
4:lung? (can’t see number 4 so assuming)
5:left bronchi
6:diaphragm
7:pharynx
8:trachea
9:right bronchi
10:bronchioles
11:alveoli

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HELP! THIS IS SCIENCE

Alexxx [7]

This is known as Rutherford's gold foil experiment. To align with J.J Thompson's Plum Pudding Model, he expects a beam of alpha particles to just pass through the gold foil undisturbed. However, some were deflected at certain angles. Alpha particles are positive, so it would just go straight through the nucleus, but will deflect if it hits the electrons. Therefore, the answer is: Particles that struck the nucleus went straight.

7 0

4 years ago

Read 2 more answers

Compound 2 contains 2.0g of hydrogen and 32.0g oxygen. What is the percent compound of each element?

Dmitriy789 [7]

Note down the formula below

$\boxed{\sf Mass\%\;of\; element=\dfrac{Mass\:of\:the\: element}{Mass\;of\:the\: compound}\times 100}$

Mass of the compound

$\\ \sf\longmapsto 32+2=34g$

Mass % of Hydrogen:-

$\\ \sf\longmapsto \dfrac{2}{34}\times 100$

$\\ \sf\longmapsto \dfrac{1}{17}\times 100$

$\\ \sf\longmapsto 5.8\%$

Mass % of Oxygen:-

$\\ \sf\longmapsto \dfrac{32}{34}\times 100$

$\\ \sf\longmapsto \dfrac{16}{17}\times 100$

$\\ \sf\longmapsto 94.2\%$

7 0

3 years ago

What is the predicted outcome for very low mass stars (those that are smaller than about the mass of the sun)?

White raven [17]

Answer:

In case of low-mass stars,the outer layers of the low mass stars are expelled as the core collapses such that the outer layers form a planetary nebula.

Explanation:

In case of low-mass stars,the outer layers of the low mass stars are expelled as the core collapses such that the outer layers form a planetary nebula. The core remains as a white dwarf and finally become a black dwarf as it cools down. A low mass star consumes its core hydrogen and turns it into helium over its lifetime.

4 0

3 years ago

The entropy change for a real, irreversible process is equal to:______.

taurus [48]

Answer:

The entropy change for a real, irreversible process is equal to zero.

The correct option is 'c'.

Explanation:

Lets look around all the given options -:

(a) the entropy change for a theoretical reversible process with the same initial and final states , since the entropy change is equal and opposite in reversible process , thus this option in not correct.

(b) equal to the entropy change for the same process performed reversibly ONLY if the process can be reversed at all. Since , the change is same as well as opposite too . Therefore , this statement is also not true .

(c) zero. This option is true because We generate more entropy in an irreversible process. Because no heat moves into or out of the surroundings during the procedure, the entropy change of the surroundings is zero.

(d) impossible to tell. This option is invalid , thus incorrect .

Hence , the correct option is 'c' that is zero.

8 0

3 years ago

Draw the bridged bromonium ion that is formed as an intermediate during the bromination of this alkene. include hydrogen atoms,

gogolik [260]

<h2>Answer</h2>

Bromination:

Any reaction or process in which bromine (and no other elements) are introduced into a molecule.

Bromonium Ion:

The bromonium ion is formed when alkenes react with bromine. When the π cloud of the alkene (acting as a nucleophile) approaches the bromine molecule (acting as an electrophile), the σ-bond electrons of Br2 are pushed away, resulting in the departure of the bromide anion.(2)

Mechanism:

Step 1:

In the first step of the reaction, a bromine molecule approaches the electron-rich alkene carbon–carbon double bond. The bromine atom closer to the bond takes on a partial positive charge as its electrons are repelled by the electrons of the double bond. The atom is electrophilic at this time and is attacked by the pi electrons of the alkene [carbon–carbon double bond]. It forms for an instant a single sigma bond to both of the carbon atoms involved (2). The bonding of bromine is special in this intermediate, due to its relatively large size compared to carbon, the bromide ion is capable of interacting with both carbons which once shared the π-bond, making a three-membered ring. The bromide ion acquires a positive formal charge. At this moment the halogen ion is called a "bromonium ion".

Step 2:

When the first bromine atom attacks the carbon–carbon π-bond, it leaves behind one of its electrons with the other bromine that it was bonded to in Br2. That other atom is now a negative bromide anion and is attracted to the slight positive charge on the carbon atoms. It is blocked from nucleophilic attack on one side of the carbon chain by the first bromine atom and can only attack from the other side. As it attacks and forms a bond with one of the carbons, the bond between the first bromine atom and the other carbon atoms breaks, leaving each carbon atom with a halogen substituent.

In this way the two halogens add in an anti addition fashion, and when the alkene is part of a cycle the dibromide adopts the trans configuration.

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3 years ago