How does the cardiovascular system help the body get rid of waste??

2 answers:

8 0

Not the rectum but the kidneys. cardiovascular system involves the blood flow and the urination to get rid of wastes. as the body had damaged blood cells or worn down cells, it moves the cells through three processes. respiration, when the cells are obtaining oxygen, then perspiration, when the blood or oxygen is used during motion or sleep, the body will emit heat and eventually move to the final stage whish it felicitation, when the body gets rid of unwanted wastes like bacteria, salts and excess liquids, through urination, the rectum moves SOLID wastes like the skin of the foods you eat, carbohydrates, and proteins that your body cannot use anymore. so your answer is partially correct but it is through facilitation.

Pani-rosa [81]2 years ago

4 0

The answer is the rectum

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$\large \boxed{\mathbf{1.36; 3.6 \times 10^{\mathbf{-9}}}\textbf{mol/L}}$

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1 .Calculate the hydronium ion concentration

We can use an ICE table to organize the calculations.

HF + H₂O ⇌ H₃O⁺ + F⁻

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C/mol·L⁻¹: -x +x +x

E/mol·L⁻¹: 2.7 - x x x

$K_{\text{a}} = \dfrac{\text{[H}_{3}\text{O}^{+}] \text{F}^{-}]} {\text{[HF]}} = 7.2 \times 10^{-4}\\\\\dfrac{x^{2}}{2.7 - x} = 7.2 \times 10^{-4}\\\\\text{Check for negligibility of }x\\\\\dfrac{2.7}{7.2 \times 10^{-4}} = 4000 > 400\\\\\therefore x \ll 2.7\\\dfrac{x^{2}}{2.7} = 7.2 \times 10^{-4}\\\\x^{2} = 2.7 \times 7.2 \times 10^{-4} = 1.94 \times 10^{-3}\\x = 0.0441\\\text{[H$_{3}$O$^{+}$]}= \text{x mol$\cdot$L$^{-1}$} = \text{0.0441 mol$\cdot$L$^{-1}$}$

2. Calculate the pH

$\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{0.0441} = \large \boxed{\mathbf{1.36}}$

3. Calculate [C₆H₅O⁻]

C₆H₅OH + H₂O ⇌ C₆H₅O⁻ + H₃O⁺

2.7 x 0.0441

$K_{\text{a}} = \dfrac{0.0441x} {2.7} = 1.6 \times 10^{-10}\\\\0.0441x = 1.6 \times 10^{-10}\\x = \dfrac{1.6 \times 10^{-10}}{0.0441} = \mathbf{3.6 \times 10^{\mathbf{-9}}}\textbf{ mol/L}\\\text{The concentration of phenoxide ion is $\large \boxed{\mathbf{3.6 \times 10^{\mathbf{-9}}}\textbf{ mol/L}}$}$