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2 years ago

A decrease of blood ph from 7.4 to 7.2 causes hemoglobin to ________.

1 answer:

5 0

Answer: A decrease of blood ph from 7.4 to 7.2 causes hemoglobin to ________. C) increase its binding of H+.

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6K + B2O3 → 3K2O + 2B

atroni [7]

Answer:

104.84 moles

Explanation:

Given data:

Moles of Boron produced = ?

Mass of B₂O₃ = 3650 g

Solution:

Chemical equation:

6K + B₂O₃ → 3K₂O + 2B

Number of moles of B₂O₃:

Number of moles = mass/ molar mass

Number of moles = 3650 g/ 69.63 g/mol

Number of moles = 52.42 mol

Now we will compare the moles of B₂O₃ with B from balance chemical equation:

B₂O₃ : B

1 : 2

52.42 : 2×52.42 = 104.84

Thus from 3650 g of B₂O₃ 104.84 moles of boron will produced.

6 0

2 years ago

Hafnium has six naturally occurring isotopes: 0.16% of 174Hf, with an atomic weight of 173.940 amu; 5.26% of 176Hf, with an atom

ser-zykov [4K]

Answer:

177.277amu

Explanation:

the total occuring isotopes for Hafnium is =6.

First isotope had an atomic weight of 173.940amu

Second isotope =175.941amu

Third isotope =176.943amu

Fourth isotope=177.944amu

Fifth isotope. =178.946amu

sixth isotope .179.947amu

Average atomic weight of Hafnium= sum of all the atomic weights of the isotopes/ Total occuring isotopes

Thus, 173.940amu+175.941amu+176.943amu+177.944amu+178.946amu+179.947amu.= 1063.661amu

Average atomic weight= 1063.661amu /6 = 177.2768333amu

= 177.277amu to 3 decimal places.

6 0

2 years ago

Consider the following reaction (X = Cl or Br) which statement s is are correct?

statuscvo [17]

A.S OLOS kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkll

5 0

2 years ago

A 0.72-mol sample of PCl5 is put into a 1.00-L vessel and heated. At equilibrium, the vessel contains 0.40 mol of PCl3(g) and 0.

Sonja [21]

Answer:

Equilibrium constant for $PCl_5$ is 0.5

Equilibrium constant for decomposition of $NO_2$ is $1.79 \times 10^{-14}$

Explanation:

$PCl_5$ dissociates as follows:

$PCl_5 \rightleftharpoons PCl_3+Cl_2$

initial 0.72 mol 0 0

at eq. 0.72 - 0.40 0.40 0.40

Expression for the equilibrium constant is as follows:

$k=\frac{[PCl_3][Cl_2]}{[PCl_5]}$

Substitute the values in the above formula to calculate equilibrium constant as follows:

$k=\frac{[0.40/1][0.40/1]}{0.32/1} \\=\frac{0.40 \times 0.40}{0.32} \\=0.5$

Therefore, equilibrium constant for $PCl_5$ is 0.5

Now calculate the equilibrium constant for decomposition of $NO_2$

It is given that $3.3 \times 10^{-3} \%$ is decomposed.

$NO_2$ decomposes as follows:

$2NO_2 \rightleftharpoons 2NO + O_2$

initial 1.0 M 0 0

at eq. concentration of $NO_2$ is:

$[NO_2]_{eq}=1-(0.000066) = 0.999934\ M$

$[NO]_{eq}=6.6 \times 10^{-5}\ M$

$[O_2]_{eq}=3.3\times 10^{-5} = 3.3\times 10^{-5}\ M$

Expression for equilibrium constant is as follows:

$K=\frac{[NO]^2[O_2]}{[NO_2]^2}$

Substitute the values in the above expression

$K=\frac{[6.6\times 10^{-5}]^2[3.3 \times 10^{-5}]}{[0.999934]^2} \\=1.79\times 10^{-14}$

Equilibrium constant for decomposition of $NO_2$ is $1.79 \times 10^{-14}$

8 0

2 years ago

the stability of atomic nuclei is related to the _____. ratio of protons to electrons ratio of neutrons to protons number of pro

jenyasd209 [6]

ratio of neutrons to protons

6 0

2 years ago

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