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sashaice [31]
3 years ago
8

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

Chemistry
2 answers:
Cloud [144]3 years ago
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]3 years ago
4 0
The answer is the rectum 
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How many grams of Co3+ are present in 2.34 grams of cobalt(III) nitrite?
Assoli18 [71]

Answer:

m_{Co^{3+}}=0.563gCo^{3+}

Explanation:

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In this case, since these mole-mass relationships are understood in terms of the moles of the atoms forming the considered compound, we first realize that the chemical formula of the cobalt (III) nitrate is Co(NO₃)₃ whereas there is a 1:1 mole ratio of the cobalt (III) ion (molar mass = 58.93 g/mol) to the entire compound. In such a way, we first compute the moles of the salt (molar mass = 58.93 g/mol) and then apply the aforementioned mole ratio to obtain the grams of the required cation:

m_{Co^{3+}}=2.34gCo(NO_3)_3*\frac{1molCo(NO_3)_3}{244.95 gCo(NO_3)_3} *\frac{1molCo^{3+}}{1molCo(NO_3)_3} *\frac{58.93gCo^{3+}}{1molCo^{3+}} \\\\m_{Co^{3+}}=0.563gCo^{3+}

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4 0
3 years ago
Which of the data sets qualitative data?
Kaylis [27]

Answer:

Uh, the second one?

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3 0
3 years ago
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What volume of a 5.00M solution of hydrochloric acid contains 8.00mol of HCl?
rjkz [21]

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7 0
3 years ago
The rate constant for a certain reaction is measured at two different temperatures:
Talja [164]

Answer: The activation energy Ea for this reaction is 22689.8 J/mol

Explanation:

According to Arrhenius equation with change in temperature, the formula is as follows.

ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]

k_1 = rate constant at temperature T_1 = 2.3\times 10^8

k_2 = rate constant at temperature T_2 = 4.8\times 10^8

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T_2 = temperature = 376.0^0C=(273+376)=649K

Putting in the values ::

ln \frac{4.8\times 10^8}{2.3\times 10^8} = \frac{-E_{a}}{8.314}[\frac{1}{649} - \frac{1}{553}]

E_a=22689.8J/mol

The activation energy Ea for this reaction is 22689.8 J/mol

3 0
3 years ago
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