Planets reflect sunlight. Most planets are close enough to the sun that they can reflect its light and that is why you can see them. Some planets that are far from the sun do not light up because they have no light to reflect.
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Answer:
The correct answer is 25 mL graduated cylinder (it should be used in all the cases)
Explanation:
In order to measure 25.00 ml sample of a solution it should be used a 25 mL graduated cylinder, as it is previously and properly calibrated. The other laboratory glassware, beaker and erlenmeyer, have graduations which are approximate, so they are used when exact volumes are not needed.
ii) graduated cylinder has the least uncertainly. It is more accurate than a beaker or erlenmeyer (to within 1%)
iii) A 25 mL graduated cylinder should be used because it is the most accurate lab glassware (between those were mentioned: beaker, erlenmeyer).
Answer:
E - Be and O
A - Mg and N
E - Li and Br
F - Ba and Cl
B - Rb and O
Explanation:
Be and O
Be is a metal that loses 2 e⁻ to form Be²⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form BeO (E-MX).
Mg and N
Mg is a metal that loses 2 e⁻ to form Mg²⁺ and N is a nonmetal that gains 3 e⁻ to form O³⁻. For the ionic compound to be neutral, it must have the form Mg₃N₂ (A-M₃X₂).
Li and Br
Li is a metal that loses 1 e⁻ to form Li⁺ and Br is a nonmetal that gains 1 e⁻ to form Br⁻. For the ionic compound to be neutral, it must have the form LiBr (E-MX).
Ba and Cl
Ba is a metal that loses 2 e⁻ to form Ba²⁺ and Cl is a nonmetal that gains 1 e⁻ to form Cl⁻. For the ionic compound to be neutral, it must have the form BaCl₂ (F-MX₂).
Rb and O
Rb is a metal that loses 1 e⁻ to form Rb⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form Rb₂O (B-M₂X).
Answer:
V = 22.41 L
Explanation:
Given data:
Mass of nitrogen = 14.0 g
Volume of gas at STP = ?
Gas constant = 0.0821 atm.L/mol.K
Solution:
Number of moles of gas:
Number of moles = mass/molar mass
Number of moles= 14 g/ 14 g/mol
Number of moles = 1 mol
Volume of gas:
PV = nRT
1 atm × V = 1 mol × 0.0821 atm.L/mol.K × 273 K
V = 22.41 atm.L / 1 atm
V = 22.41 L
