Which statement defines an ore?

[tex]1s^2 2s^2 2p^6 3s^2 3p^3[/tex] is the correct electron configuration for an element with five electrons in the third energy level.

Elements are the simplest substances which cannot be broken down using chemical methods.

The shell nearest to the nucleus, 1n, can carry two electrons, while the next shell, 2n, can carry eight, and the third shell, 3n, can carry up to eighteen.

The third shell carries 18 electrons; 2 in a 3s orbital; 6 in three 3p orbitals; and 10 in five 3d orbitals. The fourth shell carries  32 electrons; 2 in a 4s orbital; 6 in three 4p orbitals; 10 in five 4d orbitals; and 14 in seven 4f orbitals.

The element would be phosphorus. Its electron configuration is [tex]1s^2 2s^2 2p^6 3s^2 3p^3[/tex]

Learn more about elements here:

brainly.com/question/1896898

#SPJ2

Answer:

0.1828g/cm³

Explanation:

density= mass÷volume

m= 12.6g

v= 4.1×4.1×4.1 = 68.921cm³

•

density= 12.6÷68.921 = 0.1828g/cm³

Answer:

The answer is "2.78".

Explanation:

Given values:

CD= 0.44

Formula:

[tex]\bold{f_r= \frac{v^{2}}{ g(s-1)D}}[/tex]

g=9.8

s= 2.7

D= 0.06

[tex]\to f_r=\frac{2.5^2}{9.8(2.7-1)0.06}\\\\[/tex]

        [tex]=\frac{2.5 \times 2.5}{9.8 \times 1.7 \times 0.06 }\\\\=\frac{6.25}{.9996 }\\\\=6.252501[/tex]

[tex]\phi = \frac{82\times v}{ \sqrt{cD} \times f_r^{-1.5}}\\\\[/tex]

   [tex]= \frac{82 \times 0.25 }{ \sqrt{0.44} \times 6.25^{1.5}}\\\\=2.4285\\[/tex]

[tex]\frac{\bigtriangleup P f_1 s_1}{L} = \frac{\bigtriangleup Pf_w}{L}(1+\phi)\\[/tex]

           [tex]=\frac{2fwSwv^2 (1+2.4285)}{D}\\\\[/tex]

[tex]Re= \frac{D \bar v Sw}{M_w}\\[/tex]

     [tex]=\frac{0.06 \times 2.5 \times 1000 }{0.001}\\\\=\frac{150 }{0.001}\\\\= 150 \times 10^{3}\\\\= 1.50 \times 10^{5}\\\\[/tex]

[tex]fw= 0.00389[/tex]

[tex]\to \frac{\bigtriangleup P f_1 s_1}{L}[/tex]

[tex]\to \frac{2 \times 0.00389 \times 1000 \times2.5^2 \times 3.4265}{0.06}\\\\\to 2.78[/tex]

Answer: lithium- 3 protons, 3 electrons, 3 neutrons

Oxygen- nucleon no. 16, Atomic no. 8, 8 protons, overall charge (-2)

Sorry I don't the others. Hope this helps

Answer:

1.65485688 × 10121 grams

Explanation:

Aside from those discrete items, anything like rice or flour can be measured out to make up 100 grams. For example, 1/2 cup (125 ml or about 4 and 1/2 fl. oz) of dry rice weighs very close to 100 grams.

Explanation:

eam=%abudance×mass+%abudance×mass

eam=19.8%×10.013/100+80.2%×11.009/100

eam=198.2574/100+882.9218/100

eam=1081.1792/100

eam=10.811792

eam=10.812

Answer:

2.76 gram

Explanation:

2AI + 6HCl → 2AlCl3 + 3H2

Answer

Al +2 HCl - AlCl2 + H2

Answer:

35.453 g

Explanation:

one mole of chlorine is 34.453 g

Answer:

One mole of chlorine atoms has a mass of 35.453g. A mole of something means that you have 6.023×1023 units of that something. So if you have a mole of Cl atoms you have 6.023×10  

Hope this helps <3

Need thanks? Comment /hearthelp

Answer:

[tex]\large \boxed{\text{a)188.4 g; b) 98.67 $\, \%$}}[/tex]

Explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:                      98.08           392.18

             2Cr + 3H₂SO₄ ⟶ Cr₂(SO₄)₃ + 3H₂

To solve the stoichiometry problem, you must

a) Mass of Cr₂(SO₄)₃

(i) Mass of pure H₂SO₄

[tex]\text{Mass of pure} = \text{165 g impure} \times \dfrac{\text{85.67 g pure} }{\text{100 g impure}} = \text{141.36 g pure}[/tex]

(ii) Moles of H₂SO₄

[tex]\text{Moles of H$_{2}$SO}_{4} = \text{141.36 g H$_{2}$SO}_{4} \times \dfrac{\text{1 mol H$_{2}$SO}_{4}}{\text{98.08 g H$_{2}$SO}_{4}} = \text{1.441 mol H$_{2}$SO}_{4}[/tex]

(iii) Moles of Cr₂(SO₄)₃

The molar ratio is 1 mol Cr₂(SO₄)₃:3 mol H₂SO₄ [tex]\text{Moles of Cr$_{2}$(SO$_{4}$)}_{3} = \text{1.441 mol H$_{2}$SO}_{4} \times \dfrac{\text{1 mol Cr$_{2}$(SO$_{4}$)}_{3}}{\text{3 mol H$_{2}$SO}_{4}} = \text{0.4804 mol Cr$_{2}$(SO$_{4}$)}_{3}[/tex]

(iv) Mass of Cr₂(SO₄)₃ [tex]\text{Mass of Cr$_{2}$(SO$_{4}$)}_{3} = \text{0.4804 mol Cr$_{2}$(SO$_{4}$)}_{3} \times \dfrac{\text{392.18 g Cr$_{2}$(SO$_{4}$)}_{3}}{\text{1 mol Cr$_{2}$(SO$_{4}$)}_{3}} = \textbf{188.4 g Cr$_{2}$(SO$_{4}$)}_{3}\\\text{The mass of Cr$_{2}$(SO$_{4}$)$_{3}$ formed is $\large \boxed{\textbf{188.4 g}}$}[/tex]

b) Percentage yield

It is impossible to get a yield of 485.9 g. I will assume you meant 185.9 g.

[tex]\text{Percentage yield} = \dfrac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 \, \% = \dfrac{\text{185.9 g}}{\text{188.4 g}} \times 100 \, \% = \mathbf{98.67 \, \%}\\\\\text{The percentage yield is $\large \boxed{\mathbf{98.67 \, \%}}$}[/tex]