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soldier1979 [14.2K]
3 years ago
14

The nucleus of a fluorine atom has a charge of? A. +19 B. 0 C. +9 D. +1

Chemistry
1 answer:
posledela3 years ago
7 0

Answer:

A fluorine atom has nine protons and nine electrons, so it is electrically neutral. If a fluorine atom gains an electron, it becomes a fluoride ion with an electric charge of -1.

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4.1 shows a plant cell. g For Examiner's Use n. C D Fig. 4.1 (i) Name the type of plant cell shown in Fig. 4.1. [1]​
Vinil7 [7]

Answer:

palisade cell due to presence of chloroplasts

7 0
2 years ago
The second order reaction A → Products takes 13.5 s for the concentration of A to decrease from 0.740 M to 0.319 M. What is the
Ivanshal [37]

The rate constant of the second order reaction is 0.137 M-1s-1.

<h3>What is the rate constant?</h3>

For the second order reaction we can write;

1/[A] = kt + 1/[A]o

[A]o = initial concentration

[A] = final concentration

k = rate constant

t = time

Now;

1/0.319 = 13.5k + 1/ 0.740

1/0.319 - 1/0.740 = 13.5k

3.13 - 1.35 = 13k

k = 3.13 - 1.35/13

k = 0.137 M-1s-1

Learn more about second order reaction:brainly.com/question/12446045

#SPJ1

8 0
2 years ago
What occurs when materials drop out of wind or water?
pantera1 [17]
Erosion weathering decomposition one of those
7 0
3 years ago
Calculate the solubilities of the following compounds in a 0.02 M solution of barium nitrate using molar concentrations, first i
Law Incorporation [45]

Answer:

a. 1.7 × 10⁻⁴ mol·L⁻¹; b. 5.5 × 10⁻⁹ mol·L⁻¹

c. 2.3 × 10⁻⁴ mol·L⁻¹;    5.5 × 10⁻⁸ mol·L⁻¹

Explanation:

a. Silver iodate

Let s = the molar solubility.  

                     AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq); Ksp = 3.0 × 10⁻⁸

E/mol·L⁻¹:                               s               s

K_{sp} =\text{[Ag$^{+}$][IO$_{3}$$^{-}$]} = s\times s =  s^{2} = 3.0\times 10^{-8}\\s = \sqrt{3.0\times 10^{-8}} \text{ mol/L} = 1.7 \times 10^{-4} \text{ mol/L}

b. Barium sulfate

                     BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq); Ksp = 1.1 × 10⁻¹⁰

I/mol·L⁻¹:                                0.02             0

C/mol·L⁻¹:                                 +s              +s

E/mol·L⁻¹:                            0.02 + s          s

K_{sp} =\text{[Ba$^{2+}$][SO$_{4}$$^{2-}$]} = (0.02 + s) \times s \approx  0.02s = 1.1\times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{0.02} \text{ mol/L} = 5.5 \times 10^{-9} \text{ mol/L}

c. Using ionic strength and activities

(i) Calculate the ionic strength of 0.02 mol·L⁻¹ Ba(NO₃)₂

The formula for ionic strength is  

\mu = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\\mu = \dfrac{1}{2} (\text{[Ba$^{2+}$]}\cdot (2+)^{2} + \text{[NO$_{3}$$^{-}$]}\times(-1)^{2}) = \dfrac{1}{2} (\text{0.02}\times 4 + \text{0.04}\times1)= \dfrac{1}{2} (0.08 + 0.04)\\\\= \dfrac{1}{2} \times0.12 = 0.06

(ii) Silver iodate

a. Calculate the activity coefficients of the ions

\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(1)^{2}\sqrt{0.06} = -0.51\times 0.24 = -0.12\\\gamma = 10^{-0.12} = 0.75

b. Calculate the solubility

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq)

K_{sp} =\text{[Ag$^{+}$]$\gamma_{Ag^{+}}$[IO$_{3}$$^{-}$]$\gamma_{IO_{3}^{-}}$} = s\times0.75\times s \times 0.75 =0.56s^{2}= 3.0 \times 10^{-8}\\s^{2} = \dfrac{3.0 \times 10^{-8}}{0.56} = 5.3 \times 10^{-8}\\\\s =2.3 \times 10^{-4}\text{ mol/L}

(iii) Barium sulfate

a. Calculate the activity coefficients of the ions

\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(2)^{2}\sqrt{0.06} = -0.51\times16\times 0.24 = -0.50\\\gamma = 10^{-0.50} = 0.32

b. Calculate the solubility

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq

K_{sp} =\text{[Ba$^{2+}$]$\gamma_{ Ba^{2+}}$[SO$_{4}$$^{2-}$]$\gamma_{ SO_{4}^{2-}}$} = (0.02 + s) \times 0.32\times s\times 0.32 \approx  0.02\times0.10s\\2.0\times 10^{-3}s = 1.1 \times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{2.0 \times 10^{-3}} \text{ mol/L} = 5.5 \times 10^{-8} \text{ mol/L}

7 0
3 years ago
. Write the following isotope in nuclide notation (e.g., “ ”): copper-70
vekshin1

Answer:

{\boxed{\text{$_{29}^{70}${Cu}}}

Explanation:

The atomic number (Z) of copper is 29 and this isotope has an atomic mass (A) of 70.

The general symbol for an isotope E is _{Z}^{A}\text{E}.

The atomic number is a left subscript, and the atomic mass is a left superscript.

\rm {\text{The nuclide notation for copper-70 is }}{\boxed{\textbf{$_{29}^{70}${Cu}}}

4 0
3 years ago
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