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2 years ago

True or False Hypoxia results from higher oxygen levels.

Chemistry

2 answers:

OverLord2011 [107]2 years ago

8 0

Answer:

True

Explanation:

Atmospheric pressure reduces the amount of oxygen in the air, therefore leading to Hypoxia. Higher altitues contribute.

shtirl [24]2 years ago

5 0

The answer to this would be true

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D. Energy is lost as body heat by each organism

dimaraw [331]

A and D I’m not sure tho

5 0

2 years ago

What type of chemical reaction is given by the following balanced equation: Al₂O3 → AL+ O₂

rjkz [21]

Answer:

Decomposition

Explanation:

3 0

2 years ago

1. A 25 g rock is placed in a graduated cylinder with water. The volume of the liquid rises from 18.3 mL to 21.4 mL Calculate th

MaRussiya [10]

Answer:

1. $\rho=8.06g/cm^3$

2. $H=10cm$

Explanation:

Hello,

1. In this case, since the volume of the rock is obtained via the difference between the volume of the cylinder with the water and the rock and the volume of the cylinder with the water only:

$V=21.4mL-18.3mL=3.1mL$

Thus, the density turns out:

$\rho=\frac{m}{V}=\frac{25g}{3.1cm^3} \\\\\rho=8.06g/cm^3$

2. In this case, given the density and mass of aluminum we can compute its volume as follows:

$V=\frac{m}{\rho}=\frac{1080g}{2.7g/cm^3}=400cm^3$

Moreover, as the volume is also defined in terms of width, height and length:

$V=W*H*L$

The height is computed to be:

$H=\frac{V}{L*W}=\frac{400cm^3}{5cm*8cm}\\ \\H=10cm$

Best regards.

3 0

2 years ago

The hydrogen chloride (HCl) molecule has an internuclear separation of 127 pm (picometers). Assume the atomic isotopes that make

natta225 [31]

Answer:

the energy of the third excited rotational state $\mathbf{E_3 = 16.041 \ meV}$

Explanation:

Given that :

hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm

Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.

Thus; the reduced mass μ = $\dfrac{m_1 \times m_2}{m_1 + m_2}$

μ = $\dfrac{1 \times 35}{1 + 35}$

μ = $\dfrac{35}{36}$

∵ 1 μ = 1.66 × 10⁻²⁷ kg

μ = $\\ \\ \dfrac{35}{36} \times 1.66 \times 10^{-27} \ \ kg$

μ = 1.6139 × 10⁻²⁷ kg

$r_o = 127 \ pm = 127*10^{-12} \ m$

The rotational level Energy can be expressed by the equation:

$E_J = \dfrac{h^2}{8 \pi^2 I } \times J ( J +1)$

where ;

J = 3 ( i.e third excited state) &

$I = \mu r^2_o$

$E_J= \dfrac{h^2}{8 \pi \mu r^ 2 \mur_o } \times J ( J +1)$

$E_3 = \dfrac{(6.63 \times 10^{-34})^2}{8 \times \pi ^2 \times 1.6139 \times 10^{-27} \times( 127 \times 10^{-12}) ^ 2 } \times 3 ( 3 +1)$

$E_3= 2.5665 \times 10^{-21} \ J$

We know that :

1 J = $\dfrac{1}{1.6 \times 10^{-19}}eV$

$E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV$

$E_3 = 16.041 \times 10 ^{-3} \ eV$

$\mathbf{E_3 = 16.041 \ meV}$

8 0

2 years ago

What would be the major product obtained from hydroboration–oxidation of the following alkenes?

zmey [24]

Answer:

a. 3-methylbutan-2-ol

b. 2-methylcyclohexan-1-ol

Explanation:

For this reaction, we must remember that the hydroboration is an "anti-Markovnikov" reaction. This means that the "OH" will be added at the least substituted carbon of the double bond.

In the case of 2-methyl-2-butene, the double bond is between carbons 2 and 3. Carbon 2 has two bonds with two methyls and carbon 3 is attached to 1 carbon. Therefore the "OH" will be added to carbon three producing 3-methylbutan-2-ol.

For 1-methylcyclohexene, the double bond is between carbons 1 and 2. Carbon 1 is attached to two carbons (carbons 6 and 7) and carbon 2 is attached to one carbon (carbon 3). Therefore the "OH" will be added to carbon 2 producing 2-methylcyclohexan-1-ol.

See figure 1

I hope it helps!

8 0

2 years ago