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Marat540 [252]
3 years ago
13

As carbon dioxide enters systemic blood, it causes more oxygen to dissociate from hemoglobin, which in turn allows more CO2 to c

ombine with hemoglobin and more bicarbonate ions to be generated Group of answer choices True False
Chemistry
1 answer:
vagabundo [1.1K]3 years ago
5 0

Answer:

The correct answer is "False".

Explanation:

It is false that as carbon dioxide enters systemic blood, it causes more oxygen to dissociate from hemoglobin. Once an atom of oxygen binds to hemoglobin, hemoglobin change its shape and makes easier than a second and a third atom of oxygen binds towards it. This change in conformation makes no possible that carbon dioxide can cause that oxygen dissociates from hemoglobin.

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Rank the following fertilizers in decreasing order of mass percentage of nitrogen:
charle [14.2K]
<h3>Answer:</h3>

        NH₃ > NH₄NO₃ > (NH₄)₂HPO₄ > (NH₄)₂SO₄ > KNO₃ > (NH₄)H₂PO₄

<h3>Soution:</h3>

In (NH₄)₂HPO₄:

Mass of Nitrogen  =  N × 2  =  14 × 2  =  28 g.mol⁻¹

Molar Mass of (NH₄)₂HPO₄  =  132.06 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of (NH₄)₂HPO₄ × 100

Mass %age  =  28 g.mol⁻¹ / 132.06 g.mol⁻¹ × 100

Mass %age  =  21.20 %

In (NH₄)₂SO₄:

Mass of Nitrogen  =  N × 2  =  14 × 2  =  28 g.mol⁻¹

Molar Mass of (NH₄)₂SO₄  =  132.14 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of (NH₄)₂SO₄ × 100

Mass %age  =  28 g.mol⁻¹ / 132.14 g.mol⁻¹ × 100

Mass %age  =  21.18 %

In KNO₃:

Mass of Nitrogen  =  N × 1  =  14 × 1  =  14 g.mol⁻¹

Molar Mass of KNO₃  =  101.10 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of KNO₃ × 100

Mass %age  =  14 g.mol⁻¹ / 101.10 g.mol⁻¹ × 100

Mass %age  =  13.84 %

In (NH₄)H₂PO₄:

Mass of Nitrogen  =  N × 1  =  14 × 1  =  14 g.mol⁻¹

Molar Mass of (NH₄)H₂PO₄  =  115.03 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of (NH₄)H₂PO₄ × 100

Mass %age  =  14 g.mol⁻¹ / 115.03 g.mol⁻¹ × 100

Mass %age  =  12.17 %

In NH₃:

Mass of Nitrogen  =  N × 1  =  14 × 1  =  14 g.mol⁻¹

Molar Mass of NH₃  =  132.14 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of NH₃ × 100

Mass %age  =  14 g.mol⁻¹ / 17.03 g.mol⁻¹ × 100

Mass %age  =  82.20 %

In NH₄NO₃:

Mass of Nitrogen  =  N × 2  =  14 × 2  =  28 g.mol⁻¹

Molar Mass of NH₄NO₃  =  80.04 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of NH₄NO₃ × 100

Mass %age  =  28 g.mol⁻¹ / 80.04 g.mol⁻¹ × 100

Mass %age  =  34.98 %

5 0
3 years ago
11. If four gases in a cylinder each exert 6 atm, what is the total pressure exerted by the
Debora [2.8K]

Answer:

24 atm is the total pressure exerted by the gases

Explanation:

We propose this situation:

In a vessel, we have 4 gases (for example, hydrogen, Xe, methane and chlorine)

Each of the gases has the same pressure:

6 atm → hydrogen

6 atm → xenon

6 atm → methane

6 atm → chlorine

To determine the total pressure, we sum all of them:

Partial pressure H₂ + Partial pressure Xe + Partial pressure CH₄ + Partial pressure Cl₂ = Total P

6 atm + 6 atm + 6 atm + 6 atm = 24atm

8 0
3 years ago
Write an equation that shows the formation of a chromium(ii) ion from a neutral chromium atom.
daser333 [38]
The   equation  that  shows  the  formation  of  chromium (ii) ion  from  neutral  chromium  atom  is  as  follow
Cr ---> cr^2+   +  2e-
Cr^2+ is  the  chromium  ion   with  oxidation  state  of  two  which  is  one  of  the  common  ion  of  chromium.  Other  common  ion   of  chromium  include  chromium  of  oxidation  state  6  and  3

8 0
3 years ago
Read 2 more answers
Which of the following chemical equations is unbalanced? A. 2C + H2 CH4 B. 2Al2O3 4Al + 3O2 C. 2H2O2 2H2O + O2 D. 2C2H6 + 7O2 4C
Tanya [424]

Answer:

A. 2C + H₂ ⟶ CH₄  

Explanation:

A. 2C + H₂ ⟶ CH₄

UNBALANCED. 2C on the left and 1C on the right

B. 2Al₂O₃ ⟶ 4Al + 3O₂  

Balanced. Same number of each type of atom on each side.

C. 2H₂O₂ ⟶ 2H₂O + O₂  

Balanced. Same number of each type of atom on each side.

D. 2C₂H₆ + 7O₂ ⟶ 4CO₂ + 6H₂O

Balanced. Same number of each type of atom on each side.

8 0
3 years ago
What word best describes volume
alekssr [168]

Answer:

volume is the amount of space something occupies

Explanation:

7 0
3 years ago
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