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vichka [17]
3 years ago
11

Connectivity is an example of which type of property?​

Chemistry
1 answer:
aleksklad [387]3 years ago
3 0

Answer:

physical property

Explanation:

......

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Question 2 (1 point) Saved
Diano4ka-milaya [45]

Answer:

False, the electrons are on the outside protons and neutrons are in the center

Explanation:

5 0
3 years ago
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A solution of copper sulfate (CuSO4) produced by bioleaching has a concentration of 0.319 g/dm3 Relative atomic masses (Ar): Cu
Tanya [424]

moles Cu produced : 0.002

<h3>Further explanation</h3>

Concentration of copper sulfate (CuSO₄) : 0.319 g/dm³

MW CuSO₄ :

\tt =63.5+32+4\times 16=159.5~g/mol

mol CuSO₄ /dm³ :

\tt \dfrac{0.319}{159.5}=0.002

CuSO₄⇒Cu²⁺ + SO₄²⁻

mol Cu : mol CuSO₄ = 1 : 1 , so mol Cu²⁺=0.002

6 0
3 years ago
Determine the concentration of a solution made by dissolving 44.0 g of calcium chloride (CaCl2) in 0.30 L of solution. SHOW YOUR
ycow [4]

Answer:

[CaCl₂] = 1.32 M

Explanation:

We know the volume of solution → 0.30 L

We know the mass of solute → 44 g of CaCl₂

Let's convert the mass of solute to moles.

44 g . 1 mol / 110.98 g = 0.396 moles

Molarity (mol/L) → 0.396 mol / 0.3 L  = 1.32 M

6 0
3 years ago
Imagine this scenario. All of the trees and plants are removed from a steep mountainside how would the removal of the trees and
lilavasa [31]
Nothing unless it was dug out from roots if not they would grom back in a long period of time
8 0
3 years ago
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How many moles are in 3.113 g of Au?Molar mass of Au=197 g/mol
SVEN [57.7K]
<h3>Answer:</h3>

0.0157 g Au

<h3>General Formulas and Concepts:</h3>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis

<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
<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.113 g Au

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

Molar Mass of Au - 197.87 g/mol

<u>Step 3: Convert</u>

<u />3.113 \ g \ Au(\frac{1 \ mol \ Au}{197.87 \ g \ Au} ) = 0.015733 g Au

<u>Step 4: Check</u>

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

0.015733 g Au ≈ 0.0157 g Au

3 0
3 years ago
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