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neonofarm [45]
2 years ago
14

Rolls of foil are 300mm wide and 2.020mm thick. (The density of foil is 2.7 g/cm^3). What maximum length of foil can be made fro

m 1.40 kg of foil?
Chemistry
1 answer:
horrorfan [7]2 years ago
5 0

Answer:  8556 mm, or 855.6 cm (8560 mm to 3 sig figs)

Explanation:  Convert mm to cm by dividing by 10 (1cm/10mm)

Find the area of the foil face in cm^2 (30cm*0.2020cm) = 0.606 cm^2

Calculate the volume occupied by 1.40 kg of foil in cm^3.  1.40kg = 1400g

1.400g/(2.7 g/cm^3) = 518.5 cm^3 for 1.40 kg Au

   Volume = Area (of the face) * Length  

We want Length:

Length = Volume/Area

L = (518.5 cm^3/0.606 cm^2)

L = 855.6 cm (8556 mm)  Round to 3 sig figs (856 cm and 8560 mm)

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Answer:

A pure solid is heated and turns into a pure liquid.

Explanation:

No colour change recorded, only change of state, hence this is a physical change - physical changes I.e. change of state and temperature are not chemical reactions.

7 0
2 years ago
What is the electron configuration of the element with 27 protons?
adelina 88 [10]
<span>The electronic configuration of cobalt is :1s2 2s2 2p6 3s2 3p6 3d7 4s2 </span>
7 0
3 years ago
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Which of the following acts as a catalyst in your body? A. Sweat B. Enzymes C. Blood cells D. Sugar
Alenkasestr [34]
The answer is B. Enzymes.

Enzymes are biological catalysts that help cause reactions in your body.
3 0
3 years ago
What is the minimum of acetic anhydride (102.1 g/mol) required to react completely with the 2.94 grams of salicylic acid? What v
Lady bird [3.3K]

Answer:

a) the minimun of acetic anhydride required for the  reaction is 2.175 g (CH3CO)2O

b) V acetic anhydride = 2.010 mL

Explanation:

  • balanced reaction:

      C6H4OHCOOH + (CH3CO)2O  ↔  C9H8O4 + C2H4O2

⇒ mol salicylic acid = 2.94 g C6H4OHCOOH * ( mol C6H4OHCOOH / 138.121 g ) = 0.0213 mol C6H4OHCOOH

⇒ mol acetic anhydride = 0.0213 mol C6H4OHCOOH * ( mol (CH3CO)2O /  mol  C6H4OHCOOH ) = 0.0213 mol (CHECO)2O

⇒ g acetic anhydride = 0.0213 mol * ( 102.1 g/mol ) = 2.175 g CH3CO)2O

b) V = 2.175 g (CH3CO)2 * ( mL / 1.082 g ) = 2.010 mL (CH3CO)2O

4 0
2 years ago
How many moles in 4.93 x 10E23 atoms of silver?
leva [86]
<h3>Answer:</h3>

0.819 mol Ag

<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> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

4.93 × 10²³ atoms Ag

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 4.93 \cdot 10^{23} \ atoms \ Ag(\frac{1 \ mol \ Ag}{6.022 \cdot 10^{23} \ atoms \ Ag})
  2. Divide:                              \displaystyle 0.818665 \ mol \ Ag

<u>Step 4: Check</u>

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

0.818665 mol Ag ≈ 0.819 mol Ag

8 0
2 years ago
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