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3 years ago

A chemical reaction produces 11.8 moles of tin atoms how many grams of tin are made

1 answer:

7 0

Convert mols to grams by multiplying grams of tin by the number of mols.

There are 119 grams per mol

119 x 11.8 = 1404 grams

You might be interested in

Can colloidal suspensions be separated out by filtration

balu736 [363]

Answer:

Colloidal can not be separated through filtration.

Suspension can be separated through filtration.

Explanation:

Colloidal:

Colloidal consist of the particles having size between 1 - 1000 nm i.e, 0.001- 1μm. While the pore size of filter paper is 2μm. That's why we can not separate the colloidal through the filtration. However it can be separated through the ultra filtration. In ultra filtration the pore size is reduced by soaking the filter paper in gelatin and then in formaldehyde. This is only in case of when solid colloidal is present, if colloid is liquid , there is no solid particles present and ultra filtration can not be used in this case.

Suspension:

The particle size in suspension is greater than 1000 nm. The particles in suspension can be separated through the filtration. These particles are large enough and can be seen through naked eye.

5 0

3 years ago

Read 2 more answers

Newton's law stating that for every action force there is an equal and opposite reaction force; often referred to as Newton's th

Gnesinka [82]

When two bodies interact they apply forces to one another that are equal to magnitude and opposite direction

4 0

2 years ago

Consider the following reaction at a high temperature. Br2(g) ⇆ 2Br(g) When 1.35 moles of Br2 are put in a 0.780−L flask, 3.60 p

UNO [17]

Answer : The equilibrium constant $K_c$ for the reaction is, 0.1133

Explanation :

First we have to calculate the concentration of $Br_2$ .

$\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}$

$\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M$

Now we have to calculate the dissociated concentration of $Br_2$ .

The balanced equilibrium reaction is,

$Br_2(g)\rightleftharpoons 2Br(aq)$

Initial conc. 1.731 M 0

At eqm. conc. (1.731-x) (2x) M

As we are given,

The percent of dissociation of $Br_2$ = $\alpha$ = 1.2 %

So, the dissociate concentration of $Br_2$ = $C\alpha=1.731M\times \frac{1.2}{100}=0.2077M$

The value of x = 0.2077 M

Now we have to calculate the concentration of $Br_2\text{ and }Br$ at equilibrium.

Concentration of $Br_2$ = 1.731 - x = 1.731 - 0.2077 = 1.5233 M

Concentration of $Br$ = 2x = 2 × 0.2077 = 0.4154 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

$K_c=\frac{[Br]^2}{[Br_2]}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.4154)^2}{1.5233}=0.1133$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.1133

7 0

3 years ago

An(A)=25,n(B)=40,n(AuB)=50<br><br>AnB

puteri [66]

Answer:

5879

Explanation:

3896_+746864uyrhufhj

5 0

3 years ago

Look at this compound, why does magnesium has dative and covalent bonds with nitrogen? ...?

HACTEHA [7]

<span>The solid lines between N and Mg are actually ionic bonds. N has 5 valence electrons (2 of which are paired). Of the 3 that are unpaired, 2 are part of covalent bonds with adjacent carbon atoms. N accepts an extra electron to complete its octet, but gets a formal charge of -1. This allows for formation of an ionic bond with Mg, which is +2. Two of these charged N atoms therefore neutralize the charge of the central Mg. As for the coordinate (dative) covalent bonds, Mg has empty orbitals - the ionic bonds with the charged N atoms give it only 4/8 possible valence electrons.

The other two N atoms (dotted lines) have a formal charge of 0 since they form three covalent bonds with adjacent carbon atoms, but they still have a lone pair. Therefore, just to improve stability, each of these N atoms can "donate" its lone pair to Mg in order to complete its octet.

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4 0

3 years ago