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Akimi4 [234]
2 years ago
7

TRY H Vele

Chemistry
1 answer:
Strike441 [17]2 years ago
6 0
<h3>Answer:</h3>

Al- [Ne] 3s²3p¹

As- [Ar] 4s²3d¹⁰ 4p³

Explanation:

  • Electron configuration of an element shows the arrangement of electrons in the energy levels or orbitals in the atom.
  • Noble-gas configuration involves use of noble gases to write the configuration of other elements.
  • This is done by identifying the atomic number of the element and then identifying the noble gas that comes before that particular element on the periodic table.
  • For example:
  1. Aluminium: The atomic number of Al is 13. The noble gas before Aluminium is Neon which has 10 electrons. Therefore the remaining 3 electrons fills up the 3s and 3p sub orbitals.
  • Thus, the noble-gas configuration of Al is [Ne] 3s²3p¹

     2.  Arsenic, Atomic number is 33

  • Noble gas before Arsenic is Ar,. Argon has 17 electrons, then the remaining electrons fills up the 4s, 3d and 4p sub-orbitals.
  • Thus, the noble-gas configuration of As is [Ar] 4s²3d¹⁰ 4p³
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At a festival, spherical balloons with a radius of 280. Cm are to be inflated with hot air and released. The air at the festival
evablogger [386]

Answer:

8608.18 balloons

Explanation:

Hello! Let's solve this!

Data needed:

Enthalpy of propane formation: 103.85kJ / mol

Specific heat capacity of air: 1.009J · g ° C

Density of air at 100 ° C: 0.946kg / m3

Density of propane at 100 ° C: 1.440kg / m3

First we will calculate the propane heat (C3H8)

3000g * (1mol / 44g) * (103.85kJ / mol) * (1000J / 1kJ) = 7.08068 * 10 ^ 6 J

Then we can calculate the mass of the air with the heat formula

Q = mc delta T

m = Q / c delta T = (7.08068 * 10 ^ 6 J) / (1.009J / kg ° C * (100-25) ° C) =

m = 93566.96kg

We now calculate the volume of a balloon.

V = 4/3 * pi * r ^ 3 = 4/3 * 3.14 * 1.4m ^ 3 = 11.49m ^ 3

Now we calculate the mass of the balloon

mg = 0.946kg / m3 * 11.49m ^ 3 = 10.87kg

The amount of balloons is

93566.96kg / 10.87kg = 8608.18 balloons

5 0
3 years ago
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Hydrogen reacts with chlorine to form hydrogen chloride (HCl (g), Delta.Hf = –92.3 kJ/mol) according to the reaction below. Uppe
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Answer:

The enthalpy of the reaction is –184.6 kJ, and the reaction is exothermic.

Explanation:

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3 years ago
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If 15.6 grams of copper (ii) chloride react with 20.2 grams of sodium nitrate how many grams of sodium chloride can be formed? W
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Answer:

- 13.56 g of sodium chloride are theoretically yielded.

- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.

- 0.50 g of sodium nitrate remain when the reaction stops.

- 92.9 % is the percent yield.

Explanation:

Hello!

In this case, according to the question, it is possible to set up the following chemical reaction:

CuCl_2+2NaNO_3\rightarrow 2NaCl+Cu(NO_3)_2

Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:

m_{NaCl}^{by\ CuCl_2}=15.6gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2} *\frac{2molNaCl}{1molCuCl_2} *\frac{58.44gNaCl}{1molNaCl} =13.56gNaCl\\\\m_{NaCl}^{by\ NaNO_3}=20.2gNaNO_3*\frac{1molNaNO_3}{84.99gNaNO_3} *\frac{2molNaCl}{2molNaNO_3} *\frac{58.44gNaCl}{1molNaCl} =13.89gNaCl

Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:

m_{NaNO_3}^{by\ NaCl}=13.56gNaCl*\frac{1molNaCl}{58.44gNaCl}*\frac{2molNaNO_3}{2molNaCl} *\frac{84.99gNaNO_3}{1molNaNO_3}=19.72gNaNO_3

Therefore, the leftover of sodium nitrate is:

m_{NaNO_3}^{leftover}=20.2g-19.7g=0.5gNaNO_3

Finally, the percent yield is computed via:

Y=\frac{12.6g}{13.56g} *100\%\\\\Y=92.9\%

Best regards!

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