We need to know how many valence electrons are present in a given molecule in order to determine its Lewis dot structure.
<h3>
What is Lewis dot structure?</h3>
Lewis dot structure is defined as the visual representation of atoms' electrons using a diagram. It describe the bonds that exist between a molecule's lone pairs of electrons and its atoms.
A lone pair of electrons on the atom with the formal charge of 1- can typically be converted into a bonding pair that is shared with the atom that has the formal charge of 1+ in order to reduce the formal charges when they are present.
Thus, we need to know how many valence electrons are present in a given molecule in order to determine its Lewis dot structure.
To learn more about Lewis dot structure, refer to the link below:
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Answer:
1.28 atm
Explanation:
To solve this problem, you need to use the gas laws, more specifically the Combined Gas Law. It is P1V1/T1 = P2V2/T2. Simply plug your values in. But be careful! Make sure you convert your 20 degree C and 28 deg C to Kelvin, as that it the only temperature scale the Gas Laws work with. Upon plugging in your values, you get approximately 1.28 atm.
<span>Nitrogen-14 should be the correct response.</span>
Answer:
Because his first law states that an object with a net force of zero acting on it will remain at rest, if initially at rest, or it will maintain a constant velocity.
Answer:
1. 0.74mol
2. 0.42mol
3. 2.125mol
4. 0.301mol
5. 4.52 × 10^23 particles
Explanation:
Number of moles (n) in a substance can be found using the formula:
mole (n) = mass/molar mass
Using this formula, the following moles are calculated:
1. Molar of Na = 23g/mol
mole = 17/23
mole = 0.74mol
2. Molar mass of Na2SO4 = 23(2) + 32 + 16(4)
= 46 + 32 + 64
= 142g/mol
Mole = 60/142
mole = 0.42mol
3. Molar mass of CO2 = 12 + 16(2)
= 12 + 32
= 44g/mol
mole = 93.5/44
mole = 2.125mol
4. Molar mass of sodium nitrate (NaNO3) = 23 + 14 + 16(3)
= 23 + 14 + 48
= 85g/mol
mole = 25.6/85
mole = 0.301mol
5. Number of particles in one mole of a substance is 6.022 × 10^23 particles. Hence, in 0.75mol of calcium hydroxide (Ca(OH)2, there will be;
0.75mol × 6.02 × 10^23
= 4.515 × 10^23
= 4.52 × 10^23 particles
