The substances which will exist in molecular form in solution are-
A. CH₃NH₂
D. CH₃OH
The methyl amine (CH₃NH₂) and methanol (CH₃OH) is organic compounds which are hardly become ionic in nature and prefers to form as molecular state in the solution phase.
On the other hand B. LiOH and C. NH₄OH are inorganic compound and highly ionic in nature and remain as ionic state in solution.
Those are-
LiOH (s) → Li⁺(s) + OH⁻ (s)
NH₄OH→ NH₄⁺ (s) + OH⁻ (s)
1. 12 L = 12 dm³
2. 3.18 g
<h3>Further explanation</h3>
Given
1. Reaction
K₂CO₃+2HNO₃⇒ 2KNO₃+H₂O+CO₂
69 g K₂CO₃
2. 0.03 mol/L Na₂CO₃
Required
1. volume of CO₂
2. mass Na₂CO₃
Solution
1. mol K₂CO₃(MW=138 g/mol) :
= 69 : 138
= 0.5
mol ratio of K₂CO₃ : CO₂ = 1 : 1, so mol CO₂ = 0.5
Assume at RTP(25 C, 1 atm) 1 mol gas = 24 L, so volume CO₂ :
= 0.5 x 24 L
= 12 L
2. M Na₂CO₃ = 0.03 M
Volume = 1 L
mol Na₂CO₃ :
= M x V
= 0.03 x 1
= 0.03 moles
Mass Na₂CO₃(MW=106 g/mol) :
= mol x MW
= 0.03 x 106
= 3.18 g
1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N. Details about number of atoms can be found below.
How to calculate number of atoms?
The number of atoms of a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.
However, the number of moles of oxygen in glycine can be calculated using the following expression:
Molar mass of C₂H5O2N = 75.07g/mol
Mass of oxygen in glycine = 32g/mol
Hence; 32/75.07 × 7.51 = 3.2grams of oxygen in glycine
Moles of oxygen = 3.2g ÷ 16g/mol = 0.2moles
Number of atoms of oxygen = 0.2 × 6.02 × 10²³ = 1.205 × 10²³ atoms
Therefore, 1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N.
Learn more about number of atoms at: brainly.com/question/8834373
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In order to increase the strength of his electromagnet and pick up more thumbtacks Braddy should use two batteries instead of one. Using two batteries will increase the amount of current in the coils which will induce a higher electromagnetic field in the nail.
Another option would be to increase the number of coils of wire.
