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andriy [413]
3 years ago
15

The density of pure solid copper is 8.94 g/mL. What volume does 34 g of copper occupy?

Chemistry
1 answer:
patriot [66]3 years ago
4 0

hey there ! :

Density = 8.94 g/mL

mass = 34 g

Volume = ??

Therefore:

D = m / V

8.94 = 34 / V

V = 34 / 8.94

V = 3.803 mL

Hope this helps!

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If sufficient acid is used to react completely
Juli2301 [7.4K]

Solution,

Mass of Mg - 21gm

molar mass of Mg - 24gm

moles = Given mass/ Molar mass

moles = 21/24 = 0.875

1 mole of Mg produce 1 mole of H2 gas

so 0.875 mole of Mg will produce 0.875 moles of H2 gas

One mole of H2 gas = 22.4 litre

0.875 mole of H2 gas = 0.875×22.4

0.875 mole of H2 gas = 19.60 litre

So the nearest ANSWER is Option four 19.37 litre.

3 0
2 years ago
Hello people ~
Klio2033 [76]
<h3>CHEMISTRY</h3>

What is the most basic aromatic amine’s common name?

a) Benzenamine

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2 years ago
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Consider the molecular structure for linuron, an herbicide, provided in the questions below. How many pi bonds are in the molecu
Sophie [7]

In the herbicide linuron whose structure is shown in the image attached, there are four pi bonds.

A pi bond is formed by a sideways overlap of atomic orbitals. A sigma bond is formed by an end to end or head to head overlap of atomic orbitals. Pi bonds lead to the occurrence of multiple bonds in the molecule.

In the herbicide linuron whose structure is shown in the image attached to this answer, there are four pi bonds which are easily spotted as double bonds in the structure.

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6 0
3 years ago
What is Saponification!?
Tom [10]

Answer:

<h3>Saponification is a process that involves conversion of fat, oil or lipid into soap and alcohol by the action of heat in the presence of aqueous alkali. Soaps are salts of fatty acids and fatty acids are monocarboxylic acids that have long carbon chains e.g. sodium palmitate.</h3>
8 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
3 years ago
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