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snow_tiger [21]
3 years ago
5

Use the periodic table in the tools bar to answer these questions. How many moles of AgNO3 are present in 1.50 L of a 0.050 M so

lution?
Chemistry
2 answers:
IgorC [24]3 years ago
4 0
M= moles de soluto / litros de solucion 
moles de soluto = M. litros de solucion 

Moles de soluto = 0.050 M x 1.50 L = 0.075 moles de AgNo3
mixer [17]3 years ago
3 0

m = Solute moles/liters solution moles of solute =

m. Liters of solution moles of solute = 0050 m x 1.50 L = 0075 moles of AgNo3


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If you start with 5 molecules of O2 in the reaction 2Mg+O2--> 2MgO, how much Mg will you need?
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You will need 4 molecules of O2.
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Which of the following is the correct molecule for dinitrogen trisulfide?
Setler79 [48]

Answer:

A. N2S3

yep u r right

Explanation:

Dinitrogen Trisulfide N2S3 Molecular Weight -- EndMemo.

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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
In order to predict the outcome of the reaction, write the molecular, full ionic, and net ionic equations for a mixture of aqueo
Elina [12.6K]

Answer:

See explanation

Explanation:

Full molecular equation;

2NH3(aq) + AgNO3(aq) -------> [Ag(NH3)2]NO3(aq)

Full ionic equation

2NH3(aq) + Ag^+(aq) + NO3^-(aq) --------> [Ag(NH3)2]^+(aq) + NO3^-(aq)

Net ionic equation;

2NH3(aq) + Ag^+(aq) -------->  [Ag(NH3)2]^+(aq)

When Silver nitrate is mixed with a solution of aqueous ammonia, a white and cloudy solution was observed.

6 0
3 years ago
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