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Sholpan [36]
3 years ago
6

A new planet is discovered in a different solar system. The new planet is much like Earth. One difference is that the planet doe

s not have tides. What does the planet lack that Earth has?
Chemistry
2 answers:
blagie [28]3 years ago
6 0

Answer:A moon

Explanation:

Olin [163]3 years ago
5 0

Answer:

A Moon

Explanation:

I just took the test

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Which statement best describes the process that arrow 1 represents
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<span>Molecules are speeding up during boiling. Is the correct answer this question.</span>
8 0
2 years ago
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Given the two reactions PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.84×10−10, and AgCl(aq)⇌Ag+(aq)+Cl−(aq), K4 = 1.14×10−4, what is the
Nimfa-mama [501]

Answer: The value of equilibrium constant for reaction is, 1.42\times 10^{-2}

Explanation:

The given chemical equations are:

(1) PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq) ;  K_3=1.84\times 10^{-10}

(2) AgCl(aq)\rightleftharpoons Ag^{+}(aq)+Cl^-(aq) ;  K_4=1.14\times 10^{-4}

Now we have to calculate the equilibrium constant for chemical equation as:

PbCl_2(aq)+2Ag^{+}(aq)\rightleftharpoons 2AgCl(aq)+Pb^{2+}(aq) ;  K=?

We are reversing reaction 2 and multiplying reaction 2 by 2 and then adding both reaction, we get the final reaction.

The equilibrium constant for the reverse reaction will be the reciprocal of that reaction.

If the equation is multiplied by a factor of '2', the equilibrium constant of that reaction will be the square of the equilibrium constant.

If we are adding equations then the equilibrium constants will be multiplied.

The value of equilibrium constant for reaction is:

K=(\frac{1}{K_4})^2\times K_3

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

K=(\frac{1}{1.14\times 10^{-4}})^2\times (1.84\times 10^{-10})

K=1.42\times 10^{-2}

Hence, the value of equilibrium constant for reaction is, 1.42\times 10^{-2}

7 0
3 years ago
When a sample of ca(s) loses 1 mole of electrons in a reaction with a sample of o2(g) the oxygen?
Stolb23 [73]
The reaction involved in present case is:

Net Reaction: Ca   +    1/2 O2    →         CaO. ..................(1)

In terms of oxidation and reduction, the reaction can be shown at

Oxidation: Ca     →       Ca2+      +       2e- .................(2)
Reduction: 1/2O2    +   2e-      →      O2-...................(3)

From, reaction 1 it can be seen that 1 mol of Ca reacts with 1/2 mol of O2 to form 1 mol of CaO.

From, reaction 2 it can be seen that 1 mol of Ca, generates 2 mol of e-.

Thus, when 1/2 mol of Ca is used in reaction, it will lose 1 mol of electrons.
6 0
3 years ago
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