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2 years ago

What is the empirical formula

Chemistry

1 answer:

Radda [10]2 years ago

4 0

Answer:

a formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms.

Explanation:

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1 point

monitta

Answer: C) a redox reaction that produces an electric current

Explanation:

Chemical cell is a device which is used for the conversion of the chemical energy produced in a redox reaction into the electrical energy. The cell consists of the negative terminal called as anode where oxidation takes place and a positive terminal called as cathode where reduction takes place.

Electrolytic cell is a device which is used to carry out chemical reactions by the use of electrical energy. The cell consists of the negative terminal called as cathode where reduction takes place and a positive terminal called as anode where oxidation takes place.

7 0

3 years ago

Net ionic equation for Barium chloride

vaieri [72.5K]

Answer: There is one way to write it but i’ll also provide an unbalanced equation and a balanced one.

Explanation:

Unbalanced : Ba (aq) + Cl2 (aq)—-> BaCl (aq)

Balanced : 2Ba (aq) + Cl2 (aq)—> 2BaCl(aq)

3 0

3 years ago

2Mg + O2 → 2MgO<br><br> If you are burning 5.8332 g of Mg, how many grams of MgO will this make?

LekaFEV [45]

<h3>Answer:</h3>

9.6724 g MgO

<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>Stoichiometry</u>

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2Mg + O₂ → 2MgO

[Given] 5.8332 g Mg

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

[RxN] 2 mol Mg = 2 mol MgO

Molar Mass of Mg - 24.31 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of MgO - 24.31 + 16.00 = 40.31 g/mol

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 5.8332 \ g \ Mg(\frac{1 \ mol \ Mg}{24.31 \ g \ Mg})(\frac{2 \ mol \ MgO}{2 \ mol \ Mg})(\frac{40.31 \ g \ MgO}{1 \ mol \ MgO})$
Multiply/Divide: $\displaystyle 9.67241 \ g \ MgO$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 5 sig figs.</em>

9.67241 g MgO ≈ 9.6724 g MgO

5 0

3 years ago

Read 2 more answers

Can you guys help me please!!!!

Masteriza [31]

Answer:

i think it is true

Explanation:

if it correct plz plz mark as brainliest

thank you

19NBoli

4 0

3 years ago

Read 2 more answers

A cubic piece of platinum metal (specific heat capacity = 0.1256 J/°C･g) at 200.0°C is dropped into 1.00 L of deuterium oxide ('

polet [3.4K]

Answer:

$a=5.65cm$

Explanation:

Hello,

In this case, for this heat transfer process in which the heat lost by the hot platinum is gained by the cold deuterium oxide based on the equation:

$Q_{Pt}=-Q_{Deu}$

We can represent the heats in terms of mass, heat capacities and temperatures:

$m_{Pt}Cp_{Pt}(T_f-T_{Pt})=-m_{Deu}Cp_{Deu}(T_f-T_{Deu})$

Thus, we solve for the mass of platinum:

$m_{Pt}=\frac{-m_{Deu}Cp_{Deu}(T_f-T_{Deu})}{Cp_{Pt}(T_f-T_{Pt})} \\\\m_{Pt}=\frac{-1.00L*1110g/L*4.211J/(g\°C)*(41.9-25.5)\°C}{0.1256J/(g\°C)*(41.9-200.0)\°C} \\\\m_{Pt}=3860.4g$

Next, by using the density of platinum we compute the volume:

$V_{Pt}=\frac{3860.4g}{21.45g/cm^3}\\ \\V_{Pt}=180cm^3$

Which computed in terms of the edge length is:

$V=a^3$

Therefore, the edge length turns out:

$a=\sqrt[3]{180cm^3}\\ \\a=5.65cm$

Best regards.

6 0

3 years ago