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Vika [28.1K]
3 years ago
8

Pleaseee helpppppppp​

Chemistry
1 answer:
ahrayia [7]3 years ago
6 0

Answer:

Potassium

1s2 2s2 2p6 3s2 3p6 4s1

Explanation:

The atom having only one electron its outermost shell must belong to an element in group one of the periodic table.

Having noted that, we proceed to find out what element in group one that has the atom just described in the question.

That atom must belong to an element in the fourth period. The only group 1 element in the fourth period is potassium.

The electron configuration of potassium is;

1s2 2s2 2p6 3s2 3p6 4s1

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Harlamova29_29 [7]

Answer:

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Explanation:

thats what i got from my research :D

7 0
3 years ago
1.26 * 10^4 + 2.50 * 10^4 in sceintific notation
eduard
Hello! I can help you with this. First, convert them into it’s written out standard form. 10^4 is 10,000. 10,00 * 1.26 is 12,600. 10,000 * 2.5 is 25,000. 12,600 + 25,000 = 37,600 or 3.76 * 10^4 in scientific notation. The answer in scientific notation is 3.76 * 10^4.
8 0
3 years ago
At liftoff, a space shuttle uses a solid mixture of ammonium  perchlorate and aluminum powder to obtain great thrust from the vo
klemol [59]

1 - A balanced equation for the reaction is 3NH4ClO4 + 3Al -------> Al2O3 + AlCl3 + 6H2O + 3NO

2 - Ammonium perchlorate is the oxidizing agent while aluminum powder is the reducing agent.

Molar mass of aluminum oxide = 102 g/mol;

The molar mass of ammonium perchlorate =117.5 g/mol.

From the equation of reaction 3 moles of ammonium perchlorate react with 3 moles of aluminum to produce 1 mole of aluminum oxide.

That is 3 x 7.5 g of ammonium perchlorate reacts with g of aluminum to produce 102 g/mol of aluminum oxide. Also 150g.

150 g of ammonium perchlorate produces (150 * 102)/(3*117.5) g of aluminum oxide = 43.4 g of aluminum oxide.

If you need to learn more about the balanced equation, click here;

brainly.com/question/26694427

#SPJ4

5 0
2 years ago
Calculate the atomic radius, in nm, of the BCC Metal above utilizing the density and the atomic weight provided by examination b
andre [41]

Answer:

radius = 0.2195 nm

Explanation:

l = 2r

r = 0.439/2 = 0.2195

7 0
3 years ago
Suppose that 0.25 mole of gas C was added to the mixture without changing the total pressure of the mixture. How does the additi
Yanka [14]

Here we have to get the effect of addition of 0.25 moles of gas C on the mole fraction of gas A in a mixture of gas having constant pressure.

On addition of 0.25 moles of C gas, the mole fraction of gas A will be  \frac{moles of gas A}{moles of gas A + 0.25}.

The partial pressure of gas A can be written as P_{A} = x_{A}×P (where x_{A} is the mole fraction of gas A present in the mixture and P is the total pressure of the mixture.

The mole fraction of gas A in a mixture of gas A and C is = \frac{moles of gas A}{moles of gas A + moles of gas C} and \frac{moles of gas C}{moles of gas A + moles of gas C} respectively.

Thus on addition of 0.25 moles of C gas, the mole fraction of gas A will be  \frac{moles of gas A}{moles of gas A + 0.25}.

Which is different from the initial state.

4 0
3 years ago
Read 2 more answers
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