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3 years ago

Of the molecules below, only ________ is polar.a. SiCl4 b. CH4 c. CCl4 d. SeF4

Chemistry

1 answer:

ludmilkaskok [199]3 years ago

6 0

Answer:

SeF4 is a polar molecule

Explanation:

SeF4 is a polar molecule because a polar molecule is any molecule that have lone pairs of electrons in the central atom or have atoms that are electronegative and the electrons between that are covalently bonded are not evenly distributed.

The electronegative atoms of flourine in SeF4 are not evenly distributed and kind pairs of electrons are on the central atom.

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nevsk [136]

Answer:

a. $K_c = \dfrac{[ FeSCN^{3+}_{(aq)}] }{[Fe^{3+}_{(aq)}] [SCN^-_{(aq)}]}$

b. $K_p = \dfrac{[H_2]^4}{[H_2O]^4}$

Explanation:

Untuk semua jenis reaksi umum:

$aA + bB\iff cC + dD$

Konstanta kesetimbangan $K_c = \dfrac{[C]^c [D]^d}{[A]^a[B]^b}$

Dari pertanyaan yang diberikan:

a. $Fe3^+_{(aq)} + SCN^-_{ (aq)} \iff FeSCN^{3+}_{(aq) }$

Konstanta kesetimbangan:

$K_c = \dfrac{[ FeSCN^{3+}_{(aq)}] }{[Fe^{3+}_{(aq)}] [SCN^-_{(aq)}]}$

b. $3Fe_{(s)} + 4H_2O_{(g)} \iff Fe_3O_4_{(s)} + 4H_{2(g)}$

Konstanta kesetimbangan untuk tekanan parsial $K_p$

$K_p = \dfrac{[H_2]^4}{[H_2O]^4}$

Karena Fe3O4 (s) hadir sebagai padatan.

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2 years ago

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3 years ago

How many kilograms of solvent would contain 0.43 mol of CaO in a 2.5 molality solution

dalvyx [7]

<h3>Answer:</h3>

0.024 kg CaO

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Aqueous Solutions</u>

Molarity = moles of solute / liters of solution

<u>Atomic Structure</u>

Reading a Periodic Tables
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

0.41 mol CaO

2.5 M Solution

<u>Step 2: Identify Conversions</u>

1000 g = 1 kg

Molar Mass of Ca - 40.08 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CaO - 40.08 + 16.00 = 56.08 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 0.43 \ mol \ CaO(\frac{56.08 \ g \ Cao}{1 \ mol \ CaO})(\frac{1 \ kg \ CaO}{1000 \ g \ CaO})$
Multiply: $\displaystyle 0.024114 \ kg \ CaO$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs as our lowest.</em>

0.024114 kg CaO ≈ 0.024 kg CaO

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