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RSB [31]
2 years ago
10

What happens when your face is near the spoon. Please specify.​

Chemistry
1 answer:
aleksandrvk [35]2 years ago
8 0

Answer:

RBCs are disc-shaped with a flatter, concave center. This biconcave shape allows the cells to flow smoothly through the narrowest blood vessels. ... Many RBCs are wider than capillaries, but their shape provides the needed flexibility to squeeze through

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Given 3.4 grams of x compound with a molar mass of 85 g and 4.2 grams of y compound with a molar mass of 48 g How much of compou
Ulleksa [173]

Answer:

4.36~g~XY

Explanation:

In this case, we can start with the reaction:

2X + Y_2~->~2XY

If we check the reaction, we will have 2 X and Y atoms on both sides. So, <u>the reaction is balanced</u>. Now, the problem give to us two amounts of reagents. Therefore, we have to find the <u>limiting reagent</u>. The first step then is to find the moles of each compound using the <u>molar mass</u>:

3.4~g~X\frac{1~mol~X}{85~g~X}=0.04~mol~X

4.2~g~Y_2\frac{1~mol~Y_2}{48~g~Y_2}=0.0875~mol~Y_2

Now, we can <u>divide by the coefficient</u> of each compound (given by the balanced reaction):

\frac{0.04~mol~X}{1}=~0.04

\frac{0.0875~mol~Y_2}{2}=0.04375

The smallest value is for "X", therefore this is our <u>limiting reagent</u>. Now, if we use the <u>molar ratio</u> between "X" and "XY" we can calculate the moles of XY, so:

0.04~mol~X\frac{2~mol~XY}{2~mol~X}=0.04~mol~XY

Finally, with the molar mass of "XY" we can calculate the grams. Now, we know that 1 mol X = 85 g X and 1 mol Y_2 = 48 g Y_2 (therefore 1 mol Y = 24 g Y). With this in mind the <u>molar mass of XY</u> would be 85+24 = 109 g/mol. With this in mind:

0.04~mol~XY\frac{109~g~XY}{1~mol~XY}=4.36~g~XY

I hope it helps!

6 0
3 years ago
Balance the equation :
Sati [7]

The balanced equation is 2 AlI 3 ( a q ) + 3 Cl 2 ( g ) → 2 AlCl 3 ( a q ) + 3 I 2 ( g ) .

<u>Explanation:</u>

  • Aluminum has a typical oxidation condition of  3+  , and that of iodine is  1-  .   Along these lines, three iodides can bond with one aluminum. You get  AlI3.  For comparable reasons, aluminum chloride is  AlCl3.  
  • Chlorine and iodine both exist normally as diatomic components, so they are  Cl2(  g  )  also,  I2(  g  ), individually. In spite of the fact that I would anticipate that iodine should be a strong.  

Balancing the equation, we get:  

            2AlI 3(  aq  )  +  3Cl2 (  g  )  →  2AlCl3 (  aq  )   +  3 I 2  (  g  )  

  • Realizing that there were two chlorines on the left, I simply found the basic numerous of 2 and 3 to be 6, and multiplied the  AlCl  3  on the right.  
  • Normally, presently we have two  Al  on the right, so I multiplied the  AlI  3  on the left. Hence, I have 6  I  on the left, and I needed to significantly increase  I 2  on the right.  
  • We should note, however, that aluminum iodide is viciously receptive in water except if it's a hexahydrate. In this way, it's most likely the anhydrous adaptation broke down in water, and the measure of warmth created may clarify why iodine is a vaporous item, and not a strong.
3 0
3 years ago
Calculate the ph of a 0.20 m solution of iodic acid (hio3, ka = 0.17).
saveliy_v [14]
Iodic acid partially dissociates into H+ and IO3- 
Assuming that x is the concentration of H+ at equilibrium, and sine the equation says the same amount of IO3- will  be released as that of H+, its concentration is also X.  The formation of H+ and IO3- results from the loss of HIO3 so its concentration at equilibrium is 0.20 M - x
Ka = [H+] [IO3-] / [HIO3]; 
<span>Initially, [H+] ≈ [IO3-] = 0 and [HIO3] = 0.20; </span>
<span>At equilibrium [H+] ≈ [IO3-] = x and [HIO3] = 0.20 - x; </span>
<span>so 0.17 = x² / (0.20 - x); </span>
<span>Solving for x using the quadratic formula: </span>
<span>x = [H+] = 0.063 M or pH = - log [H+] = 1.2.</span>
7 0
3 years ago
Which diagram represents a gas that has been ionized
Sloan [31]
Do you have a picture of a diagram?

5 0
3 years ago
Write a balanced half-reaction describing the reduction of aqueous iron(III) cations to aqueous iron(II) cations.​
melamori03 [73]

Answer:

The answer to your question is:

Explanation:

                  Iron (III)  ⇒ Iron (II)

                  Fe⁺³ + 1e⁻   ⇒   Fe⁺²

                 

6 0
3 years ago
Read 2 more answers
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