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kiruha [24]
2 years ago
13

2.303 3-5.05x10-3 1 | - 1 T2298

Chemistry
1 answer:
bekas [8.4K]2 years ago
6 0

Answer:

230,294.9 I think it's fine just try your past or ask teacher help

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Use the periodic table to identify the element indicated by each electron configuration by typing in the
EleoNora [17]

Answer:

1s22s22p6: Neon (Ne)

1s22s22p63s23p3: Phosphorous (P)

1s22s22p63s23p64s1: Potassium (K)

1s22s22p63s23p64s2(im not sure what 308 is supposed to be): Calcium (Ca)

1s22s22p63s23p64s23d104p65s24d3: there is no pure element that ends 4d3 that I know of so this can either be Zirconium(Zr) if it ends in 4d2 or Niobium (Nb) if it ends in 4d4

Explanation:

you can look at the periodic table and the trends to find the rough idea of where the electron configuration ends, there are helpful articles and images on these, i attached an image that may help. After that you can look at the atomic number to find the number of electrons for a pure element and use the electron subshell pattern thing to find the exact number

5 0
1 year ago
Electrons orbit the nucleus like planets around the sun.​<br>true or false?
gogolik [260]

Answer: true

Explanation: Electrons orbit the nucleus of an atom the way that planets revolve around the Sun. The electrons are like the planets in the solar system. The sun is like the nucleus in the solar system. The answer to the question is true.

7 0
2 years ago
Read 2 more answers
If you have 12.5g of fluoride and 16.2g of sodium, which is the limiting reactant and how sodium fluoride in grams is your theor
Korvikt [17]

Answer:

F2 is the limiting reactant

27.6 grams of NaF is produced.

Explanation:

Balance the equation first.

2Na+ F2 ---> 2NaF

To find the limiting reactant, solve for how much NaF can be produced with Na and F2

12.5g F2 x (1 mole F2/ 38.00 grams F2)x (2 mole NaF/ 1 mole F2)

=0.658 moles NaF

16.2g Na x (1 mole Na/ 22.99 grams Na)x (2 mole NaF/ 2 mole Na)

=0.705 moles NaF

Since F2 produced the least NaF, F2 is the limiting reactant.

Now, to find how much NaF there is, use the moles solved above with F2 as the limiting reactant.

0.658 moles NaF x (41.99 grams NaF/ 1 mole NaF)= 27.6 moles NaF

27.6 moles of NaF would be theoretically produced.

8 0
3 years ago
The specific heat of liquid water is 4.184 j/g· ?c. calculate the energy required to heat 10.0 g of water from 26.5?c to 83.7?c.
Roman55 [17]
Energy required=mass*specific heat*temperature change
=10*4.184*57.2
=2393.248j
=2.39*10^3
5 0
2 years ago
Element X consist of two isotopes with masses of 62.9 and 64.9. The relative atomic mass of the element is 63.6. What is th
koban [17]

Answer:

it is 50%

Explanation:

62.9x50=3145

64.9x50=3245

3245+3145=6390

6390/100=63.9

4 0
2 years ago
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