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Zina [86]
2 years ago
8

What might happen if your respiratory center malfunctioned?

Chemistry
2 answers:
daser333 [38]2 years ago
5 0

Answer:

A) I think I might be wrong :^

Nata [24]2 years ago
3 0
The answer is letter a
You might be interested in
How many electrons are in an atom with the electron<br> configuration of 1s 2s 2p 3s!?
Viefleur [7K]

Answer:

11

Explanation:

electronic configuration = 1s^2, 2s^2, 2p^6, 3s^1

therefore total electrons = 2+2+6+1 = 11

<em>Hope</em><em> </em><em>it</em><em> </em><em>helps</em><em> </em><em>youu</em><em> </em>。◕‿◕。

6 0
2 years ago
if you react 28 grams of butene with excess hydrogen how many grams of butane would you expect to get
liberstina [14]

Answer:

The correct answer is 29 grams.

Explanation:

Based on the given question, the reaction will be,

CH3CH=CHCH3 + H2 ⇒ CH3CH2CH2CH3

The molecular weight of butene is 56 grams per mole, and the molecular weight of butane is 58 grams per mole.

Thus, it can be said that 56 grams of butene reacts with hydrogen gas and produces 58 grams of butane.

Therefore, 28 grams of butene when reacts with hydrogen gas to give,

= 58/56 * 28 = 29 grams of butane.

Hence, the mass of butane produced will be 29 grams.

8 0
2 years ago
SCIENCEEEE!!!!! HELPPPP!!!!
IRINA_888 [86]

pretty sure it's both are physical changes.

5 0
3 years ago
Read 2 more answers
How many atp molecules are produced during the citric acid cycle?
Mrrafil [7]
"Only 2 molecules" of ATP <span>produced during the citric acid cycle

Hope this helps!

</span>
7 0
3 years ago
Read 2 more answers
A container has a mixture of NO2 gas and N2O4 gas in equilibrium. The chemical reaction between the two gases is described by th
kondaur [170]

Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄

Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.

For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:

Kp = \frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }

where:

P(N₂O₄) and P(NO₂) are the partial pressure of each gas.

Calculating constant:

Kp = \frac{38.8}{61.2^{2} }

Kp = 0.0104

After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.

P(N₂O₄) + P(NO₂) = 200

P(N₂O₄) = 200 - P(NO₂)

Kp = \frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }

0.0104 = \frac{200 - P(NO_{2})  }{[P(NO_{2} )]^{2}}

0.0104[P(NO_{2} )]^{2} + P(NO_{2} ) - 200 = 0

Resolving the second degree equation:

P(NO_{2} ) = \frac{-1+\sqrt{9.32} }{0.0208}

P(NO_{2} ) = 98.7

Find partial pressure of N₂O₄:

P(N₂O₄) = 200 - P(NO₂)

P(N₂O₄) = 200 - 98.7

P(N₂O₄) = 101.3

The partial pressures are P(NO_{2} ) = 98.7 MPa and P(N₂O₄) = 101.3 MPa

3 0
3 years ago
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