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3 years ago

8

I thought heavy elements like iron & gold were dense & sank into the interior of the molten planet? How did gold get on

top of the less dense crust?

1 answer:

Pavel [41]3 years ago

6 0

Answer:

If it is denser then water it shall float but if water has a higher density the object sinks

Explanation:

You might be interested in

Why is it considered necessary to neutralize the industrial waste before allowing it to flow into the water bodies?

Aleonysh [2.5K]

Answer:

Some of the things, are extremly toxic to life

like nuculear power plants, create nuculear waste

you need to let it sit there for several years

unless if you want to get sued by the Enviromental protection agency

Explanation:

4 0

2 years ago

First person with the right answer gets brainliest thanks (btw the numbers on the right are the answers choose the right one)

kap26 [50]

Molar mass of FE2O3=2(55.85)+3(16)=159.7

2.56g*1mol/159.7*2mol/1mol*55.85g/1mol=1.79g

6 0

3 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

The two naturally occurring isotopes of antimony, 121Sb (57.21 percent) and 123Sb (42.79 percent), have masses of 120.904 and 12

Alex

Answer:

The correct answer is option c.

Explanation:

Formula used to determine an average atomic mass :

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

Mass of isotope Sb-121 = 120.904 amu

Fractional abundance of Sb-121 = 57.21% = 0.5721

Mass of isotope Sb-123 = 122.904 amu

Fractional abundance of Sb-123 = 42.79% = 0.4279

Average atomic mass of Sb:

$120.904 amu\times 0.5721+ 122.904 amu\times 0.4279=121.7598 amu \approx 121.76 amu$

7 0

2 years ago

I’m so confused somebody help lol

Marysya12 [62]

Answer:

lol☺️☺️☺️☺️☺️☺️☺️☺️

5 0

2 years ago

Read 2 more answers