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Musya8 [376]
3 years ago
5

You need 455 ml of a 75% alcohol solution. on hand, you have a 25% alcohol mixture. you also have a 90% alcohol mixture. how muc

h of each mixture will you need to add to obtain the desired solution?
Chemistry
1 answer:
marishachu [46]3 years ago
8 0
X ml - <span>25% alcohol mixture
y ml - </span><span>90% alcohol mixture

x+y = 455 

0.25x ml alcohol in </span>x ml of  25% alcohol mixture
0.9y ml alcohol in y ml of  90% alcohol mixture
0.75*455= 341.25 ml alcohol in 455 ml of  75% alcohol mixture

0.25x+0.9y=341.25 

System of equations:

x+y = 455 /*(-0.25) ------> -0.25x-0.25y = -0.25*455 
0.25x+0.9y=341.25

-0.25x-0.25y=-113.75
0.25x+0.9y=341.25    Add both equations

0.25x+0.9y-0.25x-0.25y=341.25-113.75
0.65y =227.5
y=227.5/0.65 = 350 ml  of  90% alcohol mixture

x+y=455
x+355=455
x=100 ml of  25% alcohol mixture

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6 0
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When an ionic compound such as sodium chloride (NaCl) is placed in water, the component atoms of the NaCl crystal dissociate int
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Explanation:

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b) <u>For covalent compounds</u>:

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The difference is a factor which is the number of particles resulting from the dissociation or ionization of one mole of the ionic compound.

So, calling M the molarity, you can write:

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This table show the calculations for the four solutions from the list of choices:

Compound    kind         Particles in solution  Molarity   # of particles

                                       (dissociation)              (M)          in 1 liter

A) NaCl          ionic            ions Na⁺ and Cl⁻        1.0            1.0 × 1 × 2 = 2

B) NaCl          ionic            ions Na⁺ anc Cl⁻        0.5           0.5 × 1 × 2 = 1

C) Glucose    covalent     molecules                   0.5          0.5 × 1 × 1 = 0.5

D) Glucose    covalent     molecules                   1.0           1.0 × 1  × 1 = 1

Therefore, the rank in increasing number of particles is for the list of solutions given is: C < B = D < A, which means that the solution expected to contain the greatest number of solute particles is the solution A) 1 L of 1.0 M NaCl.

8 0
3 years ago
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