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aksik [14]
3 years ago
8

Describe the flow of blood to the human body, including through each of the four chambers of the heart. Explain how the blood ch

anges as it moves through the various locations in the body.
Chemistry
2 answers:
patriot [66]3 years ago
7 0
This Is how it works blood<span> enters the </span>heart through<span> two large veins, the inferior and superior vena cava, emptying oxygen-poor </span>blood<span> from the </span>body<span> into the right atrium of the </span>heart<span>. As the atrium contracts, </span>blood flows<span> from your right atrium into your right ventricle </span>through<span> the open tricuspid valve.

Hope this helps <3</span>
xeze [42]3 years ago
3 0
Hey there,

The blood needs to flow at a great and constant speed. It can not flow very slow because that can be warning's of a heart attack because your blood is using alot of force and its trying to provide as much blood as possible to arrive at the heart in order for your body to live. This is why we should drink water because as we drink water, it helps our blood flow through out our body well and so that we ma not have any problem now and in the future.

~Jurgen 
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A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipita
mestny [16]

Answer:

The answer is:

(a) NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)

(b) NaCl

(c) 0.211 g

Explanation:

Given:

The mass of NaCl,

= 0.0860 g

The molar mass of NaCl,

= 58.44 g/mol

The volume of AgNO_3,

= 30.0 ml

or,

= 0.030 L

Molarity of AgNO_3,

= 0.050 M

Moles of NaCl will be:

= \frac{Given \ mass}{Molar \ mass}

= \frac{0.0860}{58.44}

= 0.00147 \ mol

now,

Moles of AgNO_3 will be:

= Molarity\times Volume

= 0.050\times 0.030

=0.0015 \ mol

(a)

The reaction is:

⇒ NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)

(b)

1 mole of NaCl react with,

= 1 mol of AgNO_3

0.0015 mol AgNO_3 needs,

= 0.00150 \ mol \ NaCl

Available mol of NaCl < needed amount of NaCl

So,

The limiting reagent is "NaCl".

(c)

The precipitate formed,

= 0.00147\times \frac{1}{1}\times \frac{143.32}{1}

= 0.211 \ g \ AgCl

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