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3 years ago

Look at the periodic table.

Chemistry

1 answer:

zhuklara [117]3 years ago

5 0

Answer:

$Mg^{+2}$

Explanation:

The oxidation state of S is (-2) and that of Mg is (+2) since Mg loses two electrons to S

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How many protons, neutrons, and electrons make up an ion of

yKpoI14uk [10]

Answer:

protons=15

electron=15

neutron=16

6 0

2 years ago

Cobalt (Co) has an atomic mass of 59 and an atomic number of 27. Which statement correctly describes an atom of cobalt?

marysya [2.9K]

Answer:

Explanation:

27 us the number of protons when we subtract 27 from 59 it will be 32 because a mass number is the sum of proton and neutron but tell u more on the comment

3 0

2 years ago

How many atoms of cobalt are in 4 moles of cobalt?

Anna [14]

<h3>Answer:</h3>

2 × 10²⁴ atoms Co

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

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

4 mol Co (Cobalt)

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

Avogadro's Number

<u>Step 3: Convert</u>

Set up: $\displaystyle 4 \ mol \ Co(\frac{6.022 \cdot 10^{23} \ atoms \ Co}{1 \ mol \ Co})$
Multiply/Divide: $\displaystyle 2.4088 \cdot 10^{24} \ atoms \ Co$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

2.4088 × 10²⁴ atoms Co ≈ 2 × 10²⁴ atoms Co

4 0

2 years ago

Which of these statements is true ?

uysha [10]

Answer:

I believe it's C

Explanation:

KHDUDCM

King Henry Died Unexpectedly Drinking Chocolate Milk

cL is C and L is U so you go back two decimal points on 6000. That is 60 so you have 60 L and 6 L and 60L > 6L

7 0

2 years ago

If a temperature increase from 10.0 ∘C to 22.0 ∘C doubles the rate constant for a reaction, what is the value of the activation

maria [59]

The activation energy barrier is 40.1 kJ·mol⁻¹

Use the Arrhenius equation

$\ln( \frac{k_2 }{k_1 }) = (\frac{E_{a} }{R })(\frac{ 1}{T_1} - \frac{1 }{T_2 })\\$

$\ln( \frac{2k }{k}) = (\frac{E_{a} }{\text{8.314 J} \cdot \text{K}^{-1} \text{mol}^{-1} })(\frac{ 1}{\text{283.15 K}} - \frac{1 }{\text{295.15 K }})\\$

$\ln2 = (\frac{ E_{a} }{\text{8.314 J} \cdot \text{K}^{-1} \text{mol}^{-1}}) \times 1.436 \times10^{-4}\\$

$\ln2 = E_{a} \times 1.727 \times 10^{-5} \text{ mol} \cdot \text{J}^{-1}$

$E_{a} = \frac{\ln2 }{ 1.727 \times10^{-5}\text{ mol} \cdot \text{J}^{-1}}\\$

$E_{a} = \text{40 100 J}\cdot\text{mol}^{-1} = \textbf{40.1 kJ}\cdot \textbf{mol}^{-1}$

3 0

3 years ago