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Ivan
2 years ago
6

The general area of the atom (outside the nucleus) where the e- are located is the — and —

Chemistry
1 answer:
aliya0001 [1]2 years ago
3 0
Shells, electron levels, and electron cloud all fit here.
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Determine the volume of 12.3 grams of formaldehyde gas at STP?
postnew [5]
9.184 liters CH2O at STP

I think this is correct. Good luck
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3 years ago
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What is a atomic bond ​
zalisa [80]
1.Electrons can be transferred from one atom to another.
2.Electrons can be shared between neighbouring atoms.
3.Electrons can be shared with all atoms in a material.
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3 years ago
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How many molecules of zinc oxide are there in a 2 kg sample?
Novosadov [1.4K]

There are 1.48 × 10²⁵ molecules of zinc oxide in a 2 kg sample. Details about number of molecules can be found below.

<h3>How to calculate number of molecules?</h3>

The number of molecules of a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.

According to this question, there are 2000g of ZnO in a sample. Zinc oxide has a molar mass of 81.38 g/mol.

no of moles = 2000g ÷ 81.38g/mol

no of moles = 24.57mol

number of molecules = 24.57 × 6.02 × 10²³

number of molecules = 147.95 × 10²³

Therefore, there are 1.48 × 10²⁵ molecules of zinc oxide in a 2 kg sample.

Learn more about number of molecules at: brainly.com/question/11815186

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6 0
2 years ago
The specific heat capacity of a pure substance can be found by dividing the heat needed to change the temperature of a sample of
kupik [55]

Answer:

See below

Explanation:

ΔQ = m c T        ΔQ = heat required(J)    m = mass (g)    T = C° temp change

                             c = heat capacity in J/g-C

4 0
2 years ago
Problem PageQuestion Gaseous ethane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . If of w
Nata [24]

Answer:

<h2> = ( 1.08 / 2.2 ) 100% = 49%</h2>

Explanation:

Balanced Equation: 2CH₃CH₃(g) + 5O₂(g) → 2CO₂(g) + 6H₂O(g)

Calculate moles of CH₃CH₃ and O₂

1.2 ₃₃ ( 1 ₃₃/ 30.0694 ₃₃ ) = 0.040 ₃₃

8.6 ₂ ( 1 2/ 31.998 ₂ ) = 0.27 ₃₃

Find limiting reagent  0.040 ₃₃ ( 5 ₂/ 2 ₃₃ ) = 0.10 ₂

CH₃CH₃ is the limiting Reagent

CH₃CH₃ (L.R.) O₂ CO₂ H₂O

Initial (mol) 0.040 0.27 0 0

Change (mol) -2x=-0 -5x= -0.10  +2x=+0.040 +6x=+0.12

Final (mol) 0 0.117 0.040 0.12

0.040 − 2 = 0 = 0.020

Determine percent yield

0.12 ₂ ( 18.0148 ₂ /1 ₂ ) = 2.2 ₂  

= ( 1.08 / 2.2 ) 100% = 49%

5 0
3 years ago
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