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Brilliant_brown [7]
3 years ago
7

Select the correct answer. What effect does an increase in products have on the reaction rate of a mixture at equilibrium?

Chemistry
1 answer:
zepelin [54]3 years ago
8 0

Answer:

D

Explanation:

If you add more to the product side (left), the reaction will reverse to the reactant (right) to balance out to achieve equilibrium.

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HELP IM TIMEDDDDDDDDDDDD
nadya68 [22]

Answer:

ethier a dessert or a plains

though plains can get rain in the summer

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it seems more like a dessert

3 0
3 years ago
Read 2 more answers
Disponemos de 514,5 g de ácido sulfúrico (H2SO4
EleoNora [17]

Answer:

es la d

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Explanation:

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8 0
3 years ago
Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 5.90 moles of magnesium perchlorate, Mg(ClO4)2.
Lady bird [3.3K]

Answer:

5.90, 11.8, 47.2

Explanation:

Let’s remove the parentheses and write the formula as MgCl₂O₈.

We see that 1 mol Mg(ClO₄)₂ contains 1 mol Mg atoms, 2 mol Cl atoms, and 8 mol O atoms.

∴ \text{Moles of Mg atoms} = \text{5.90 mol Mg(ClO}_{4})_{2} \times \frac{\text{1 mol Mg atoms}} {\text{1 mol Mg(ClO}_{4})_{2}} = \text{5.90 mol Mg atoms}  

\text{Moles of Cl atoms} = \text{5.90 mol Mg(ClO}_{4})_{2} \times \frac{\text{2 mol Cl atoms}} {\text{1 mol Mg(ClO}_{4})_{2}} = \text{11.8 mol Cl atoms}

\text{Moles of O atoms} = \text{5.90 mol Mg(ClO}_{4})_{2} \times \frac{\text{8 mol O atoms}} {\text{1 mol Mg(ClO}_{4})_{2}} = \text{47.2 mol O atoms}

∴ Mg, Cl, O = 5.90, 11.8, 47.2

4 0
4 years ago
If arranged in order of increasing atomic mass, which element would come after chlorine?
Eva8 [605]
Look at the periodic table the element next at the right of Cl is the Ar and the atomic mass is 39.948 is bigger than the Cl 35.45. All the elements after that has mases bigger than Cl.
3 0
4 years ago
Read 2 more answers
The Lyman series results from excited state hydrogen atoms transiting to
Nutka1998 [239]

Answer:

I can't draw diagrams on this web site but I can do with numbers I think. So an electron is moved from n = 1 to n = 5. I'm assuming I've interpreted the problem correctly; if not you will need to make a correction. I'm assuming that you know the electron in the n = 1 state is the ground state so the 4th exited state moves it to the n = 5 level.

n = 5 4th excited state

n = 4 3rd excited state

n = 3 2nd excited state

n = 2 1st excited state

n = 1 ground state

Here are the possible spectral lines.

n = 5 to 4, n = 5 to 3, n = 5 to 2, n = 5 to 1 or 4 lines.

n = 4 to 3, 4 to 2, 4 to 1 = 3 lines

n = 3 to 2, 3 to 1 = 2 lines

n = 2 to 1 = 1 line. Add 'em up. I get 10.

b. The Lyman series is from whatever to n = 1. Count the above that end in n = 1.

c.The E for any level is -21.8E-19 Joules/n^2

To find the E for any transition (delta E) take E for upper n and subtract from the E for the lower n and that gives you delta E for the transition.

So for n = 5 to n = 1, use -Efor 5 -(-Efor 1) = + something which I'll leave for you. You could convert that to wavelength in meters with delta E = hc/wavelength. You might want to try it for the Balmer series (n ending in n = 2). I think the red line is about 650 nm.

Explanation:

8 0
3 years ago
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