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klio [65]
3 years ago
12

There are on average 43 g of sugar and 355 mL can of soda please calculate the molarity of sugar in the can of soda the molar ma

ss of sugar is 342.2 g/mol
Chemistry
1 answer:
vfiekz [6]3 years ago
8 0

Explanation:

Given :

Amount of solute - sucrose (C12H22O11) = 41 g

Amount of solvent -soda  = 355-mL

Molarity of the solution with respect to sucrose= ?

Molarity(M) is a unit of concentration measuring the number of moles of a solute per liter of solution. The SI unit of molarity is mol/L.

Formula to find the molarity of solution :

               Molarity =  

Amount of solvent is given in mL, let’s convert to L :

               1 L = 1000 mL

Therefore, 355 mL in L will be :

                   

               = 0.355 L

We have the amount of solute in g, let’s calculate the number of moles first :

       Number of moles (n) =  

Molar mass of C12H22O11 = 342.29 g/mol.

Therefore, n =  

               = 0.119 moles.

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Identify whether each species functions as a Brønsted-Lowry acid or a Brønsted-Lowry base in this net ionic equation. ClO- + H2P
MaRussiya [10]

Answer:

ClO- = bronsted Lowry base

H2PO4- =Bronsted Lowry base

Explanation:

ClO- = is bronsted Lowry base because it accepts a proton in the reaction

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8 0
4 years ago
There are two naturally occurring isotopes of bromine. 79Br has a mass of 78.9183 amu. 81Br has a mass of 80.9163 amu. Determine
Ede4ka [16]
Atomic mass of Br = 79.904 
<span>Now  lets say  y%  is abundance of 79Br. </span>
<span>Then abundance of 81Br = (100 - y) </span>
<span>mass due to 79Br = 78.9183 * y/100 = 0.789183y
</span><span>mass due to 81Br = 80.9163 x (100 - y)/100 = 0.809163(100 - y) </span>
<span>Therfore</span>
<span>0.789183y+ 0.809163(100 - y) = 79.904 </span>
<span>0.789183y + 80.9163 - 0.809163y = 79.904 </span>
<span> - 0.01998y= 79.904 - 80.9163
 = - 1.0123 </span>
<span>y = 1.0123/0.01998 = 50.67% </span>
<span> 79Br = 50.67% </span>
<span>now
 81Br = 100 - 50.67 = 49.33%
hope this helps</span>
4 0
4 years ago
Which of these are equal to 3.01 x 1023 particles 76.9 g of diatomic iodine (I2) 79.9 g of diatomic bromine (Br2) 6 g of carbon
solmaris [256]
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79.9 g Br2 (1 mol / 159.81 g)</span>(6.022x10^23 particles / 1 mol ) = 3.011x10^23<span>

6 g C (1 mol / 12 g)</span>(6.022x10^23 particles / 1 mol ) = 3.011x10^23<span>

13.01 g CH4 ( 1 mol / 16.04 g )</span>(6.022x10^23 particles / 1 mol ) = 4.88x10^23 particles

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3 0
3 years ago
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IgorC [24]

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Explanation:

4 0
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3 0
3 years ago
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