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Ganezh [65]
3 years ago
7

How many grams of glucose are needed to prepare 400. g of a 2.00% (m/m) glucose solution g?

Chemistry
1 answer:
Dominik [7]3 years ago
5 0
The grams   of glucose  are  needed  to  prepare  400g  of  a 2.00%(m/m)  glucose  solution  g  is  calculated  as  below

=% m/m =mass  of the solute/mass  of  the  solution  x100

let mass of   solute  be represented  by  y
mass  of solution = 400 g
 % (m/m) = 2% = 2/100

 grams  of  glucose  is  therefore =2/100 =  y/400
by cross  multiplication

100y = 800
divide   both side  by  100

y= 8.0 grams



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What is the amount of solute required if the solution is 50 ml and the solvent is 35 ml. Solve and explain
e-lub [12.9K]

Answer:

15 mL of the solute

Explanation:

From the question given above, the following data were obtained:

Solution = 50 mL

Solvent = 35 mL

Solute =?

Solution is simply defined as:

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2 years ago
Consider the equation:
Dmitry [639]

Answer:1. Rate=k[CHCl_3]^1[Cl_2]^\frac{1}{2}

2. The rate constant (k) for the reaction is 3.50M^\frac{-1}{2}s^{-1}

Explanation:

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x = order with respect to CHCl_3

y = order with respect to Cl_2

n = x+y= Total order

1. a) From trial 1: 0.0035=k[0.010]^x[0.010]^y  (1)

From trial 2: 0.0069=k[0.020]^x[0.010]^y   (2)

Dividing 2 by 1 :\frac{0.0069}{0.035}=\frac{k[0.020]^x[0.010]^y}{k[0.010]^x[0.010]^y}

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b)  From trial 2: 0.0069=k[0.020]^x[0.010]^y   (3)

From trial 3: 0.0098=k[0.020]^x[0.020]^y   (4)

Dividing 4 by 3:\frac{0.0098}{0.0069}=\frac{k[0.020]^x[0.020]^y}{k[0.020]^x[0.010]^y}

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0.0035=k[0.010]^1[0.010]^\frac{1}{2}  

k=3.50M^\frac{-1}{2}s^{-1}

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