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Rina8888 [55]
3 years ago
10

What metirial is this

Chemistry
2 answers:
BaLLatris [955]3 years ago
7 0
The pic is very unclear sorry
aleksandrvk [35]3 years ago
3 0
I'm not sure based on the picture, but if this is chewing gum I think it would be the material of chicle. (Sorry if this didn't help)
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Why is zinc not extracted from ZnO through reduction using CO?​
Jet001 [13]
The standard Gibbs free energy of formation of ZnO from Zn is lower than that of CO2 from CO. Therefore, CO cannot reduce ZnO to Zn. Hence, Zn is not extracted from ZnO through reduction using CO
6 0
1 year ago
How many moles are in 8.63 x 103 atoms of Li?
Ira Lisetskai [31]
<h3>Answer:</h3>

1.43 × 10⁻²⁰ mol Li

<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>

  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>Explanation:</h3>

<u>Step 1: Define</u>

8.63 × 10³ atoms Li

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 8.63 \cdot 10^3 \ atoms \ Li(\frac{1 \ mol \ Li}{6.022 \cdot 10^{23} \ atoms \ Li})
  2. Multiply/Divide:                \displaystyle 1.43355 \cdot 10^{-20} \ moles \ Li

<u>Step 4: Check</u>

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

1.43355 × 10⁻²⁰ mol Li ≈ 1.43 × 10⁻²⁰ mol Li

4 0
3 years ago
Which increases the likelihood that a dead organisms will be fossilized
zzz [600]
When an organism is buried quickly there is less decay and better the chance for it to be persevere. The hard parts of the organism such as bones, shells, and teeth have a better chance of becoming fossils that softer parts of the organism. HARD BONES.
6 0
2 years ago
Is 13.0 mv at 25 °c. calculate the concentration of the zn2 (aq) ion at the cathode?
Andrews [41]
Answer: 
 Zn =⇒ Zn+2(0.10) + 2e- (anode)
 Zn+2(?M) + 2e- === Zn(s) (cathode)
 
 Zn + Zn+2(?M) ===⇒ Zn+2(0.10) + Zn 
 E = E^o -0.0592 log Q; in this case E^o is zero. 
 E = - 0.0592 /n logQ where n is the number of electrons transferred, in this
case n = 2 
 23 mV x 1 volt/1000mv = 0.023 Volts 
 0.023 = -0.0592 / 2 log(0.10) / [Zn+2]
 0.023 = -0.0296 { log 0.10 – log [Zn+2] }
 0.023 = -0.0296{ -1 - log[Zn+2] }
 0.023 = +0.0296 + 0.0296log[Zn+2]
 -0.0066 = 0.0296log[Zn+2]
 -0.22= log[Zn+2]
 [Zn+2] = 10^-0.22 = 0.603 Molar

6 0
3 years ago
Which concept does the diagram show?
salantis [7]

Answer:

temperature

Explanation:

sorry if wrong

4 0
2 years ago
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