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just olya [345]
3 years ago
14

The molar mass of glucose is 180 grams per mole (g/mol). Which of the following procedures should you carry out to make a 1 M so

lution of glucose? In 0.8 liter (L) of water, dissolve ________.
Chemistry
1 answer:
Strike441 [17]3 years ago
5 0

Answer:

You need to dissolve 144 gr of glucose in 0.8 liters of water to have a 1 M solution.

Explanation:

Molarity is the amount of moles per liter of solution --> moles/liters. We are trying to get a 1 molar solution and we also know the liters (0,8L) so

1\frac {moles-of-glucose}{liters-of-water} =\frac{X moles-of-glucose}{0.8 liters-of-water} \\

1\frac{Moles-of-glucose}{liters-of-water } * 0.8 liters of water

after canceling liters of water you get 0,8 moles of glucose.

180 \frac{grams-of-glucose}{mol-of-glucose}*0.8 mol- of- glucose=144 grams-of-glucose

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Why is the classification species not considered a group? (1 point)
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6 0
1 year ago
You take three compounds consisting of two elements and decompose them. To determine the relative masses of X, Y, and Z, you col
miss Akunina [59]

Answer:

a) LAW OF MULTIPLE PROPORTIONS

b) 0.095g, 0.71g, 0.285g respectively

c) X2Y, YZ15, X6Y

d) hence mass of compound X = 21 x 0.045 = 0.95g

mass of compound Y = 21 x 0.955 = 20.05g

Explanation:

a) The assumptions made in solving this questions is the application of the LAW OF MULTIPLE PROPORTIONS. The Law of multiple proportions states that if two elements A and B combine together to form more than one compound, then the several masses of A which chemically combine with a fixed mass of B is in a simple ratio.

for example, copper forms two oxides ; copper(I) oxide (CuO) and copper(ii) oxide(Cu2O), it is possible for the two samples of the oxides to be reduced to Cu by reacting with Hydrogen gas. as such, certain masses of oxygen combine separately with a fixed mass of Cu. then the ratios of Cu are then determined.

b) To calculate the relative masses, we take note of the three compounds given, they all have some amount of Y in them, hence we can use Y  as our relative mass, this implies that the relative mass of Y = 1g

mass of X = 0.4g

mass of Y = 4.2g

amount of X in 1g of Y = 0.4 x 1 /4.2

= 0.095g

for compound 2;

mass of Y = 1.4g

mass of Z = 1.0g

amount of Z in 1g of Y =1.0 x 1 /1.4

= 0.71g

for compound 3;

mass of X = 2.0g

mass of Y = 7.0g

amount of X in 1g of Y = 1 x 2/7

= 0.285g

c) Applying the law of multiple proportions; since elements X and Z combine with a fixed mass of Y, they must bear a simple ratio;

compound 1/compound 3 = 0.095/0.285

= 1/3

compound 1/compound 2 = 0.095/0.71

= 2/15

compound 2/ compound 3 = 0.71/0.285

= 5/2

formular for compound 1 = X2Y

formula for compound 2 = YZ15

formular for compound 3 = X6Y

d) from the formular X2Y, we can get the amount of each product in XY using the ratios

%of compound XY in X = mass of compound X / total Mass

= 0.2/4.4 = 4.5%

as such in a 21g of compound XY, %of compound Y = 1 - %of compound X = 95.5%

hence mass of compound X = 21 x 0.045 = 0.95g

mass of compound Y = 21 x 0.955 = 20.05g

5 0
3 years ago
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