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3 years ago

Relative Dating Using Index Fossils

Chemistry

1 answer:

3241004551 [841]3 years ago

3 0

Index fossils can help geologists compare rock Layers at Distance locations.

Index fossils help identify where Unconformities may have occurred.

Then that layer can be Compared to other locations.

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Transition metals often form ions with a charge of _____-

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+1 for transition metals

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3 years ago

Which type of clouds are feathery,high in the sky,and are made mostly of ice crystals?

Temka [501]

The cloud with those characteristics is #1

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3 years ago

Read 2 more answers

Relative formula mass of glucose? (C6H12O6)

Rom4ik [11]

To find the mass of glucose, you must multiply the atomic weight of each of the elements in the molecule by the subscripts in the formula:

$C_{6}=6*(12.01g)=72.06g$

$H_{12}=12*(1.008g)=12.096g$

$O_{6}=6*(15.999g)=95.994g$

Then you add all of them together:

$72.06g+12.096g+95.994g=180.15g$

Therefore, the molar weight of glucose is 180.15 grams.

3 0

3 years ago

The equilibrium constant, Kc, for the following reaction is 0.967 at 650 K. 2NH3(g) N2(g) 3H2(g) When a sufficiently large sampl

AlekseyPX

Answer: Concentration of $NH_3$ in the equilibrium mixture is 0.31 M

Explanation:

Equilibrium concentration of $H_2$ = 0.729 M

The given balanced equilibrium reaction is,

$2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$

Initial conc. x 0 0

At eqm. conc. (x-2y) M (y) M (3y) M

The expression for equilibrium constant for this reaction will be:

3y = 0.729 M

y = 0.243 M

$K_c=\frac{[y]\times [3y]^3}{[x-2y]^2}$

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

$K_c=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}$

$0.967=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}$

$x=0.80$

concentration of $NH_3$ in the equilibrium mixture = $0.80-2\times 0.243=0.31$

Thus concentration of $NH_3$ in the equilibrium mixture is 0.31 M

3 0

3 years ago

Help in chem please!!!!!!

IgorLugansk [536]

This is a one-step unit analysis problem. Since we are staying in moles, grams of our compound, and thus molar mass, is not needed.

1 mole is equal to 6.022x10²³ particles as given, so:

$1.5x10^{24} particles FeO_{2} (\frac{1mol}{6.022x10^{23}particles } )=2.49 molFeO_{2}$

<h3>Answer:</h3>

2.49 mol

Let me know if you have any questions.

3 0

3 years ago