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Elis [28]
2 years ago
5

Classify the below solids as amorphous or crystalline. emerald glass MgCl, rubber​

Chemistry
2 answers:
Stella [2.4K]2 years ago
7 0

Answer:Classify the below solids as amorphous or crystalline.

✔ crystalline

✔ amorphous

✔ crystalline

✔ amorphous

Explanation:

svet-max [94.6K]2 years ago
6 0

Answer:

Here yo

Explanation:

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Question<br> What is the molarity of a 400 mL solution containing 0.60 moles of NaCl?
Mumz [18]

Answer:

0.24 M

Explanation:

Molarity = Moles solute / Liters solution

Step 1: Identify variables

400 mL = Liters solution

0.60 moles = Moles solute

Step 2: Identify conversions

1 L = 1000 mL

Step 3: Convert mL to L

400mL(1 L/1000mL) = 0.4 L

Step 4: Find molarity

M = (0.4 L)(0.60 mol) = 0.24 M

6 0
2 years ago
Definition of instantaneous speed
Sindrei [870]

Answer:

Explanation:

The instantaneous speed is the speed of an object at a particular moment in time. And if you include the direction with that speed, you get the instantaneous velocity.

3 0
2 years ago
Lead(II) nitrate, Pb(NO3)2, and potassium iodide, KI<br><br><br> net ionic equation:
MissTica

Answer:

Pb²⁺ (aq) + 2I⁻ (aq) → PbI₂ (s)

General Formulas and Concepts:

  • Solubility Rules
  • Reaction Prediction

Explanation:

<u>Step 1: RxN</u>

Pb(NO₃)₂ (aq) + KI (aq) → PbI₂ (s) + KNO₃ (aq)

<u>Step 2: Balance RxN</u>

Pb(NO₃)₂ (aq) + 2KI (aq) → PbI₂ (s) + 2KNO₃ (aq)

<u>Step 3: Ionic Equations</u>

Total Ionic Equation:

Pb²⁺ (aq) + 2NO₃⁻ (aq) + 2K⁺ (aq) + 2I⁻ (aq) → PbI₂ (s) + 2K⁺ (aq) + 2NO₃⁻ (aq)

<em>Cancel out spectator ions.</em>

Net Ionic Equation:

Pb²⁺ (aq) + 2I⁻ (aq) → PbI₂ (s)

7 0
3 years ago
Which two atoms represent isotopes of the same element
inysia [295]
D & e represent the same element
6 0
3 years ago
How many molecules of ammonia are contained in 10.4 moles of ammonia, NH3?
Crank

Answer:

The answer is "\bold{6.26 \times 10^{24} \ molecules \ NH_3}\\\\"

Explanation:

1-mole molecules as per Avogadro number= 6.02 \times 10^{23} \ molecules\\\\

1-mole NH_3 \to 6.02 \times 10^{23} \ molecules\\\\

10.4 \ moles \ of \ \ NH_3 :

\to \frac{ 6.02 \times 10^{23} \ molecules }{ 1 \ mole\ NH_3}  \times 10.4 \ mole \ NH_3\\\\ \to 6.02 \times 10^{23} \ molecules   \times 10.4\\\\\to 62.608 \times 10^{23} \ molecules \\\\\to 6.26 \times 10^{24} \ molecules \ NH_3\\\\

6 0
2 years ago
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