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2 years ago

For a given reaction, adding a catalyst increases the rate of the reaction by

Chemistry

2 answers:

Ksju [112]2 years ago

8 0

Answer : Option 2) providing an alternate reaction pathway that has a lower activation energy

Explanation : When a catalyst is added to a chemical reaction it increases the rate of reaction by lowering the energy of activation. It provides an alternate pathway for a reaction to proceed with a lower Ea energy. Usually a catalyst never gets consumed in the reaction process. It will appear in the reaction mechanism steps but not in the overall reaction like reactants and products.

Andrej [43]2 years ago

6 0

The answer is (2) providing an alternate reaction pathway that has a lower activation energy. The lower activation energy will lead to the increase of rate of the reaction.

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One of the most recognizable corrosion reactions is the rusting of iron. rust is caused by iron reacting with oxygen gas in the

zloy xaker [14]

Answer: $Fe(s) + 6H_2O(l) + 3O_2(g) -> 4Fe(OH)_3(s)$

Explanation:

The Chemical equation for the formation of rust is:

Iron + Water + Oxygen ----> Rust

4 Fe(s) + 6 H2O(l) + 3 O2(g) → 4 Fe(OH)3(s)

The Iron Hydroxide However dehydrates to produce Fe2O3 * nH2O

3 0

2 years ago

Read 2 more answers

The breaking buffer that we use this week contains 10mM Tris, pH 8.0, 150mM NaCl. The elution buffer is breaking buffer that als

olasank [31]

Answer:

A breakdown of the breaking buffer was first listed with its respective component and their corresponding value; then a table was made for the stock concentrations in which the volume that is being added was determined by using the formula $M_1*V_1 = M_2*V_2$ . It was the addition of these volumes altogether that make up the 0.25 L (i.e 250 mL) with water

Explanation:

Given data includes:

Tris= 10mM

pH = 8.0

NaCl = 150 mM

Imidazole = 300 mM

In order to make 0.25 L solution buffer ; i.e (250 mL); we have the following component.

Stock Concentration Volume to be Final Concentration

added

1 M Tris 2.5 mL 10 mM

5 M NaCl 7.5 mL 150 mM

1 M Imidazole 75 mL 300 mM

$M_1*V_1 = M_2*V_2$ . is the formula that is used to determine the corresponding volume that is added for each stock concentration

The stock concentration of Tris ( 1 M ) is as follows:

$M_1*V_1 = M_2*V_2$ .

$1*V_1= 0.01 M *250mL\\V_1 = 2.5mL$

The stock concentration of NaCl (5 M ) is as follows:

$M_1*V_1 = M_2*V_2$ .

$1*V_1= 0.15 M *250mL\\V_1 = 7.5mL$

The stock concentration of Imidazole (1 M ) is as follows:

$M_1*V_1 = M_2*V_2$ .

$1*V_1= 0.03 M *250mL\\V_1 = 75mL$

Hence, it is the addition of all the volumes altogether that make up 0.25L (i.e 250 mL) with water.

5 0

2 years ago

How many moles of gold, Au, are in 3.60 x 10^-5 g of gold?

zlopas [31]

<h3>Answer:</h3>

1.83 × 10⁻⁷ mol Au

<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>Atomic Structure</u>

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.60 × 10⁻⁵ g Au (Gold)

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

Molar Mass of Au - 196.97 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 3.60 \cdot 10^{-5} \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au})$
Multiply: $\displaystyle 1.82769 \cdot 10^{-7} \ mol \ Au$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.82769 × 10⁻⁷ mol Au ≈ 1.83 × 10⁻⁷ mol Au

4 0

2 years ago

Judging and evaluating advertisements requires critical thinking because

Tom [10]

Judging and evaluating advertisements requires critical thinking because advertisements can contain confusing, misleading information.

3 0

2 years ago

Read 2 more answers

Water flows over Niagara Falls at the average rate of 2,400,000 kg/s, and the average height of the falls is about 50 m. Knowing

Andre45 [30]

Answer:

Power, $P=1.176\times 10^9\ W$

No of bulbs = 78400000

Explanation:

We have,

Water flows over Niagara Falls at the average rate of 2,400,000 kg/s, it mean it is mass per unit time i.e. m/t.

It falls from a height of 50 m

The gravitational potential energy of falling water is given by :

P = mgh

Power is equal to the work done divided by time taken. So,

$P=\dfrac{W}{t}\\\\P=\dfrac{mgh}{t}\\\\P=\dfrac{m}{t}\times gh$

So,

$P=2400000\times 9.8\times 50\\\\P=1.176\times 10^9\ W$

Let there are n bulbs that could power 15 W LED. It can be calculated by dividing the power by 15. So,

$n=\dfrac{1.176\times 10^9}{15}\\\\n=78400000\ \text{bulbs}$

It means that the number of bulbs are 78400000.

3 0

2 years ago