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3 years ago

7

The electron configurations of five different elements are shown below. which of these elements is expected to have the largest

second ionization energy (ie2 )?*

1 answer:

ipn [44]3 years ago

7 0

Missing question:

(a) 1s2 2s2 2p6 3s1
(b) 1s2 2s2 2p6 3s2
(c) 1s2 2s2 2p6 3s2 3p1

(d) 1s2 2s2 2p6 3s2 3p4

(e) 1s2 2s2 2p6 3s2 3p5

Answer is: a) 1s²2s²2p⁶3s¹ (sodium).

Sodium have the largest second ionization energy, because when he lost one electron(first ionization energy), he have stable electron configuration of noble gas neon (1s²2s²2p⁶), so sodium do not need to lost second electron, because he will have unstable electron configuration.

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Which of the following representations shows the correct placement of trophic levels?

OleMash [197]

You have to start listing from the bottom :

3. Secondary Consumers

2. Primary Consumers

1. Producer

4 0

3 years ago

1. If there are 100 navy beans, 27 pinto beans and 173 blackeyed peas in a container, what is the percent abundance in the conta

kompoz [17]

The answers to the two questions are:

1. The percent abundance in the container which has <u>100 navy</u>, <u>27 pinto</u>, and <u>173 black-eyed peas beans</u> is 33.3%, 9.0%, and 57.7% for navy bean, pinto bean, and black-eyed <em>peas </em>beans, respectively.

2. The <em>weighted </em>average score for the scores of 85, 75, 96 obtained from the evaluations of exams (20%), labs (75%), and homework (96%) is 84.1.

1. The percent abundance by type of bean is given by:

$\% = \frac{n}{n_{t}} \times 100$ (1)

Where:

n: is the number of each type of beans

$n_{t}$ : is the <em>total number</em> of <em>beans </em>

The <em>total number </em>of <em>beans</em> can be calculated by adding the number of all the types of beans:

$n_{t} = n_{n} + n_{p} + n_{b}$ (2)

Where:

$n_{n}$ : is the number of navy beans = 100

$n_{p}$ : is the number of pinto beans = 27

$n_{b}$ : is the number of black-eyed <em>peas </em>beans = 173

Hence, the total number of beans is (eq 2):

$n_{t} = 100 + 27 + 173 = 300$

Now, the <em>percent abundance</em> by type of bean is (eq 1):

Navy beans

$\%_{n} = \frac{100}{300} \times 100 = 33.3 \%$

Pinto beans

$\%_{p} = \frac{27}{300} \times 100 = 9.0 \%$

Black-eyed peas beans

$\%_{b} = \frac{173}{300} \times 100 = 57.7 \%$

Hence, the percent abundance by type of bean is 33.3%, 9.0%, and 57.7% for navy bean, pinto bean, and black-eyed peas beans, respectively.

2. The average score (S) can be calculated as follows:

$S = e*\%_{e} + l*\%_{l} + h*\%_{h}$ (3)

Where:

e: is the score for exams = 85

l: is the score for lab reports = 75

h: is the score for homework = 96

$%_{e}$ $\%_{e}$ : is the <em>percent</em> for<em> exams</em> = 70.0%

$\%_{l}$ : is the <em>percent </em>for <em>lab reports</em> = 20.0%

$\%_{h}$ : is the <em>percent </em>for <em>homework </em>= 10.0%

Then, the <u>average score</u> is:

$S = 85*0.70 + 75*0.20 + 96*0.10 = 84.1$

We can see that if the score for each evaluation is 100, after multiplying every evaluation for its respective percent, the final average score would be 100.

Therefore, the <em>weighted </em>average score will be 84.1.

Find more about percents here:

brainly.com/question/255442?referrer=searchResults

brainly.com/question/22444616?referrer=searchResults

I hope it helps you!

4 0

3 years ago

Calculate the percent composition of muscovite mica. Its chemical

Vikki [24]

Answer:

Explanation:

(KF)2(Al2O3)3(SiO2)6(H2O)

molecular formula

K₂F₂Al₆Si₆O₂₂H₂

2 x 39 + 2 x 19 + 6 x 27 + 6 x 28 + 22 x 16 + 2 x 2

= 78 + 38 + 162+ 168+ 352+ 4

= 802

Percentage of K = 78 x 100 / 802

= 9.72 %

Percentage of F = 38 x 100 / 802

= 4.74 %

Percentage of Al = 162 x 100 / 802

= 20.2 %

Percentage of Si = 168 x 100 / 802

= 20.9 %

Percentage of O = 352 x 100 / 802

= 43.9 %

Percentage of H = 4 x 100 / 802

= .54 %

7 0

4 years ago

DUE TODAY EXTREMELY IMPORTANT PLEASE PLEASE PLEASE HELP ME ASAPPPPPPPP CHEMISTRY!!!!!!!!!!!!!!!!!!! I WOULD APPRECIATE IT ALOT T

musickatia [10]

Answer:

Here's what I get

Explanation:

3. Molar concentration by formula.

$\begin{array}{rcl}M_{\text{a}}V_{a} & = & M_{\text{b}}V_{b}\\M_{a} \times \text{0.025 00 L} & = & \text{0.3840 mol/L} \times \text{0.034 52 L}\\0.025 00M_{a}\text{ L} & = & \text{0.013 26 mol}\\M_{a}&= &\dfrac{\text{0.013 26 mol}}{\text{0.025 00 L}}\\\\& = &\textbf{0.5302 mol/L}\end{array}$

(i) Comparison of molar concentrations

The formula gives a calculated value of 0.5302 mol·L⁻¹.

Dimensional analysis gives a calculated value of 0.1767 mol·L⁻¹.

The first value is three times the second.

It is wrong because the formula assumes that the acid supplies just enough moles of H⁺ to neutralize the OH⁻ from the NaOH.

Instead, I mol of H₃PO₄ provides 3 mol of H⁺, so your calculated concentration is three times the true value.

(ii) When is the formula acceptable?

The formula is acceptable only when the molar ratio of acid to base is 1:1.

Examples are

HCl + NaOH ⟶ NaCl + H₂O

H₂SO₄ + Ca(OH)₂ ⟶ CaSO₄ + 2H₂O

H₃PO₄ + Al(OH)₃ ⟶ AlPO₄ + 3H₂O

8 0

3 years ago

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . What is the theore

Masja [62]

Answer:

71g (assuming experimental data)

Explanation:

The balanced equation for this reaction:

$H_{2} SO_{4}(aq)$ + $2NaOH (s)$ → $Na_{2}SO_{4} (l)$ + $2H_{2} O (l)$

Molar mass of H2SO4 = 98.1 g/mol

molar mass of NaOH = 40g/mol

Molar mass of Na2SO4 = 142.04g/mol

⇒ 1 mole or 98.1g of H2SO4 will yield 1 ole of NaSO4; alternately, 2 moles or 49 ×2 = 80g of NaOH produces 1 mole of NaSO4.

<em>Therefore, limiting reactant is NaOH.</em>

Assuming actual experiment is 20g of NaOH,

1 mole - 40g

x moles - 20g = $\frac{20}{40}$ = 0.5 moles

⇒1 mole of Na2SO4 - 142.04g

∴ 0.5 moles = 142.04 × 0.5

<u>= 71.02g</u>

7 0

3 years ago