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shepuryov [24]
2 years ago
12

Most living things need oxygen to survive. These two body systems bring oxygen into your body and then move to all your body par

ts. What are these two body systems? A) respiratory and muscular B) digestive and circulatory C) respiratory and digestive D) circulatory and respiratory
Chemistry
2 answers:
77julia77 [94]2 years ago
8 0
The answer is <span>D) circulatory and respiratory</span>
Murljashka [212]2 years ago
7 0
Your answer would be D. As the circulatory system pumps oxygen rich blood through the body, and the respiratory system helps you to breath. Hope this helps!
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≡↔∛ππΔ≅×π⊄∴<br>khvjh j hjgvvvvvvvvvvvvvh jh
cupoosta [38]

its 40

I just took the test right now to get this answer

7 0
3 years ago
Help PlS AFAP and thank you so much
Triss [41]

Answer:

B

Explanation:

idk how to explain, B is the definition of conduction

8 0
3 years ago
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staci is attending a music festival. the ticket to the festival costs $87.96, Staci pleans to purchase $30.00 t-shirts from the
ziro4ka [17]

The total amount that Staci had in hand is $200.00.

The ticket at that time costs $87.96.

One Tshirt costs, $30.00.

Now that she has paid ticket, she will deduct the ticket from the cash she has in hand, so as to know the amount she is remaining with.

Tha is, $200-$87.96 = $112.04.

∴ In hand Staci has $112.04 extra cash whereby she can purchase more t shirts.

To get the number of tshirt she can buy, she will divide the amount she has in hand by the amount per piece of t shirt.

For example, $112.04 / $30 = $3.73.

∴ Staci can only purchase threet shirts only at a total cost of $90 for all the t shirts which is $30 × 3 = $90.

Staci has excess of 22.04 after deducting ticket and the other amount she has used to purchase the extra t shirts.

That is, $112.04 - $90 = $22.04.

3 0
3 years ago
Read 2 more answers
Can someone please help????nThis is sooooo hard
Rashid [163]

Answer:

See explanation.

Explanation:

I highly suggest you watch OChem Tutor's videos on IUPAC nomenclature because the actual naming would take a lot of time to teach in text-based format. But here is how to name them:

1) I think there are two seperate pictures for number 1. The molecule on the left is 1-pentene and the one on the right is 4-methyl-1-pentene. If the whole thing is one molecule but there is just a bond missing where the red marker numbers are, that molecule would be 9-methyl-1,6-decadiene.

2) 4-methyl-2-pentene

3) 2,4-octadiene

4) 1,5-nonadiene

5) 2,5-dimethyl-3-hexene

6) 3,6-dimethyl-2,4-heptadiene

7) 2,5,5-trimethyl-2-hexene

6 0
2 years ago
Sodium phosphate is added to a solution that contains 0.0030 M aluminum nitrate and 0.016 M calcium chloride. The concentration
gavmur [86]

Explanation:

It is given that aluminium nitrate and calcium chloride are mixed together with sodium phosphate.

And, K_{sp} of AlPO_{4} = 9.84 \times 10^{-21}

        K_{sp} of Ca_{3}(PO_{4})_{2} = 2.0 \times 10^{-29}

Let us assume that the solubility be "s". And, the reaction equation is as follows.

        AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}

     9.84 \times 10^{-21} = s \times s

             s = 9.92 \times 10^{-11}

Also,     Ca_{3}(PO_{4})_{2} \rightleftharpoons 3Ca^{2+} + 2PO^{3-}_{4}

                2 \times 10^{-29} = (3s)^{3} \times (2s)^{2}

                            s = 7.14 \times 10^{-7}

This means that first, aluminium phosphate will precipitate.

Now, we will calculate the concentration of phosphate when calcium phosphate starts to precipitate out using the K_{sp} expression as follows.

         K_{sp} = [Ca^{2+}]^{3}[PO^{3-}_{4}]^{2}

          2.0 \times 10^{-29} = (0.016)^{3}[PO^{3-}_{4}]^{2}

       2.0 \times 10^{-29} = 4.096 \times 10^{-6} \times [PO^{3-}_{4}]^{2}

       [PO^{3-}_{4}]^{2} = 4.88 \times 10^{-24}

                             = 2.21 \times 10^{-12} M

Similarly, calculate the concentration of aluminium at this concentration of phosphate as follows.

             AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}

           K_{sp} = [Al^{3+}][PO^{3-}_{4}]

       9.84 \times 10^{-21} = [Al^{3+}] \times 2.21 \times 10^{-12}

                [Al^{3+}] = 4.45 \times 10^{-9} M

Thus, we can conclude that concentration of aluminium will be 4.45 \times 10^{-9} M when calcium begins to precipitate.

7 0
2 years ago
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