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aev [14]
3 years ago
7

The instrument used to measure the volume of a liquid is the...

Chemistry
1 answer:
hjlf3 years ago
8 0

Answer:

Chemists use beakers, flasks, burets and pipets to measure the volume of liquids.

Explanation:

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Who helps recycle nutrients through an ecosystem? A. producers and decomposers B. producers and consumers C. consumers and decom
s2008m [1.1K]

D. Producers im sure of it hope im right

8 0
3 years ago
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Give two examples of an ionic compound. Chemical name and chemical structure.
Papessa [141]
NaCl (Sodium chloride)
LiF (Lithium fluoride)
4 0
3 years ago
What is the mass of 2.9 moles of calcium? Explain please
Alik [6]

Answer:

116 g

Explanation:

From the question given above, the following data were obtained:

Number of mole of calcium = 2.9 moles

Mass of calcium =.?

The mole and mass of a substance are related according to the following formula:

Mole = mass / molar mass

With the above formula, we can obtain the mass of calcium. This can be obtained as follow:

Number of mole of calcium = 2.9 moles

Molar mass of calcium = 40 g/mol

Mass of calcium =.?

Mole = mass / molar mass

2.9 = mass of calcium / 40

Cross multiply

Mass of calcium = 2.9 × 40

Mass of calcium = 116 g

Therefore, the mass of 2.9 moles of calcium is 116 g.

4 0
3 years ago
The pKb of the base cyclohexamine, C6H11NH2, is 3.36. What is the pKa of the conjugate acid, C6H11NH3
bonufazy [111]

Answer:

10.64

Explanation:

Let's consider the basic reaction of cyclohexamine, C₆H₁₁NH₂.

C₆H₁₁NH₂(aq) + H₂O(l) ⇄ C₆H₁₁NH₃⁺(aq) + OH⁻     pKb = 3.36

C₆H₁₁NH₃⁺ is its conjugate acid, since it donates H⁺ to form C₆H₁₁NH₂. C₆H₁₁NH₃⁺ acid reaction is as follows:

C₆H₁₁NH₃⁺(aq) + H₂O(l) ⇄ C₆H₁₁NH₂(aq) + H₃O⁺(aq)   pKa

We can find the pKa of C₆H₁₁NH₃⁺ using the following expression.

pKa + pKb = 14.00

pKa = 14.00 - pKb = 14.00 - 3.36 = 10.64

6 0
3 years ago
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
2 years ago
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