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2 years ago

in diagram of the heart, the right side is colored blue to indicate its low in oxygen and high in ____________

Chemistry

2 answers:

marissa [1.9K]2 years ago

7 0

Answer:

<h2>CO2 (Carbon dioxide)</h2>

Explanation:

The right side is collored blue to indicate that the blood pumping through here is low on oxygen, but high in carbon dioxide. Because the blood is so low on oxygen, and high in carbon dioxide (a byproduct of Cellular Respiration), the blood will be pumped to the lungs where the CO2 is removed, and oxygen is taken in.

larisa86 [58]2 years ago

7 0

High in Co2 (carbon dioxide)

Normally the pressure on the right side of the heart and in the pulmonary arteries is lower than the pressure on the left side of the heart and in the aorta. This is because: the right side of the heart pumps blue (deoxygenated – little or no oxygen) blood returning from the body back to the lungs.

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What is the atomic number of the as yet undiscovered element in which the 8s and 8p electron energy levels?

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The atomic number of the undiscovered element is 168

Element 118 will have just filled its 7p orbitals. therefore the predicted element to fill completely up to its 8 p orbital would have to filled a whole set of s, p, d, f and g orbitals

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10 months ago

URGENT CHEMISTRY EXPERT!

vovangra [49]

Answer:

Part 1: 7.42 mL; Part 2: 3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ 2Cu₃(PO₄)₂(s)

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Part 1. Volume of reactant

(a) Balanced chemical equation.

$\rm 2Na_{3}PO_{4} + 3CuCl_{2} \longrightarrow Cu_{3}(PO_{4})_{2} + 6NaCl$

(b) Moles of CuCl₂

$\text{Moles of CuCl}_{2} =\text{ 16.7 mL CuCl}_{2} \times \dfrac{\text{0.200 mmol CCl}_{2}}{\text{1 mL CuCl}_{2}} = \text{3.340 mmol CuCl}_{2}$

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The molar ratio is 2 mmol Na₃PO₄:3 mmol CuCl₂

$\text{Moles of Na$_{3}$PO}_{4} = \text{3.340 mmol CuCl}_{2} \times \dfrac{\text{2 mmol Na$_{3}$PO}_{4}}{\text{3 mmol CuCl}_{2}} =\text{2.227 mmol Na$_{3}$PO}_{4}$

(d) Volume of Na₃PO₄

$V = \text{2.227 mmol Na$_{3}$PO}_{4}\times \dfrac{\text{1 mL Na$_{3}$PO}_{4}}{\text{0.300 mmol Na$_{3}$PO}_{4}} = \text{7.42 mL Na$_{3}$PO}_{4} \\\\\text{The reaction requires $\large \boxed{\textbf{7.42 mL Na$_{3}$PO}_{4}}$}$

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You write molecular formulas for the solids, and you write the soluble ionic substances as ions.

According to the solubility rules, metal phosphates are insoluble.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)

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To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.

The net ionic equation is

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