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kondaur [170]
3 years ago
11

2. What happens to the reaction rate if a catalyst is added?

Chemistry
2 answers:
Ghella [55]3 years ago
6 0

Answer:

A catalyst speeds up a chemical reaction

Explanation:

It increases the reaction rate by lowering the activation energy for a reaction.

mrs_skeptik [129]3 years ago
6 0
It will speed up the chemical reaction.
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aluminum has a specific heat of 0.897 j/g°C. how much heat is needed to raise the temperature of a 79 gram piece of aluminum 28°
frutty [35]

Answer:

Approximately 2000 J.

General Formulas and Concepts:

<u>Thermodynamics</u>

Specific Heat Formula: q = mcΔT

  • <em>q</em> is heat (in J)
  • <em>m</em> is mass (in g)
  • <em>c</em> is specific heat (in J/g °C)
  • ΔT is change in temperature (in °C or K)

Explanation:

<u>Step 1: Define</u>

<em>Identify variables</em>

[Given] <em>c</em> = 0.897 J/g °C

[Given] <em>m</em> = 79 g

[Given] ΔT = 28°C

[Solve] <em>q</em>

<em />

<u>Step 2: Solve for </u><em><u>q</u></em>

  1. Substitute in variables [Specific Heat Formula]:                                             q = (79 g)(0.897 J/g °C)(28 °C)
  2. Multiply [Cancel out units]:                                                                               q = (70.863 J/°C)(28 °C)
  3. Multiply [Cancel out units]:                                                                               q = 1984.16 J

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs as our lowest.</em>

1984.16 J ≈ 2000 J

6 0
2 years ago
Gold has a density of about 20.0g/mL this means that each milliliter has a mass of about __ grams.
qwelly [4]

Answer:

20.0 grams

Explanation:

If the density of gold is 20.0 g/mL, then we can multiply it by 1 mLto find the weight of 1 mL of gold.

20.0\frac{grams}{mL}*1mL=20.0 grams

8 0
3 years ago
Read 2 more answers
an ecologist who is studying the mineral requirements and daily water intake of a plant species is studying
vampirchik [111]
They are studying the planet's ecosystem.
6 0
3 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

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. 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>

  1. Set up:                              \displaystyle 3.60 \cdot 10^{-5} \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au})
  2. 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
Cl2(g) + 2kBr(s) ---&gt; 2KCl(s) + Br2(g) Rewrite the equation and write the color of each chemical under its name. 2. What type
kolezko [41]

Answer:

Cl2(g) (green/yellow mix) + 2KBr(s) (white) ---> 2KCl(s) (violet) + Br2(g) (reddish brown)

This chemical reaction is a redox type.

Explanation:

Look at the oxidation state, when the number increase your element gets oxidated, when the number decrease, the elements it's getting reduced.

8 0
2 years ago
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