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denis23 [38]
2 years ago
14

Do you guys know any cool app for revision apart from quiz let

Chemistry
1 answer:
hammer [34]2 years ago
7 0
Kahoot, there are some pretty great reviews on there but you may need to do some searching
You might be interested in
Please help me I need these answers
Cerrena [4.2K]

1 is true

2 is d 7

3 is

            1             e

            2             b

            3             d

            4             a

            5             c

8 0
2 years ago
Within the main group of elements, what do elements in a family have in common?
statuscvo [17]
They have same chemical properties
for example the elements of group VIII are called noble gases .They are unreactive, odorless and colorless.
8 0
2 years ago
How many elements are in Li2SO4?how many elements are in li2s 04 how many elements are in li2s 04 ​
Free_Kalibri [48]

Answer:

3

Explanation:

the three elements involved in this compound are Li, S, O.

lithium, sulfur, and oxygen. Which create the ionic compound, "Lithium Sulfate."

7 atoms total, since there are two lithium, four oxygen, and one sulfate atom. this is a white inorganic salt.

6 0
2 years ago
Why are covalent bonds between hydrogen and nitrogen or oxygen polar? see section 2.1 ( page 57) ?
ASHA 777 [7]
Polar molecules exhibit an unequal balance of charges between the individual elements of the compound. This is brought about by the large difference in their electronegativities. The H atom has the least amount of electronegativity. Then, it is a known periodic trend, that as you go downwards in a group, electronegativity decreases, and increase as you go from left to right. Thus, you can deduce that the most electronegative elements are found in the upper right corner which includes O, N and F atoms. Any bond created between Hydrogen and any of O, N and F atoms is a polar bond.
3 0
3 years ago
2NO (g) + O2 (g) →2NO2 (g) At equilibrium [NO] = 2.4 × 10 -3 M, [O2] = 1.4 × 10 -4 M, and [NO2] = 0.95 M.
azamat

Answer:

K=1.12x10^9

Explanation:

Hello there!

Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:

K=\frac{[NO_2]^2}{[NO]^2[O_2]}

Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.

Best regards!

4 0
3 years ago
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