The addition of a catalyst in a chemical reaction

1 answer:

6 0

The addition of catalyst in a chemical reaction has 2 functions
1. if you add negative catalyst then the rate of chemical reaction decreases
2. if you add positive catalyst then the rate of chemical reaction increases.

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Nitrogen is an element. What is the smallest particle of nitrogen that has all the properties of nitrogen?

oksian1 [2.3K]

Answer:

The structure of the atom.

Explanation:

The correct answer is - The structure of the atom.

Reason -

An atom is the smallest unit of matter that retains all of the chemical properties of an element.

6 0

3 years ago

At a temperature of 280 K, the gas in a cylinder has a volume of 20.0 liters. If the volume of the gas is decreased to 10.0 lite

seraphim [82]

To solve this we assume that the gas inside is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant pressure and number of moles of the gas the ratio T/V is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

T1 / V1 = T2 / V2

T2 = T1 x V2 / V1

T2 = 280 x 20.0 / 10

8 0

2 years ago

Read 2 more answers

Can someone pls help, I’m sorry if it’s a lot

Helga [31]

Answer: 2738

Explaining: you gotta add

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: $\displaystyle 9.67241 \ g \ MgO$