Answer:
Explanation:
element - a basic substance that can't be simplified (hydrogen, oxygen, gold, etc...) molecule - two or more atoms that are chemically joined together (H2, O2, H2O, C6H12O6, etc...) compound - a substance that contains more than one element (H2O, C6H12O6, etc...)
Answer:
Explanation:
Two moles of magnesium (Mg) and five moles of oxygen (O2) are placed in a reaction vessel. When magnesium is ignited, it reacts with oxygen. What is the limiting reactant in this experiment?
Mg + O2 → MgO (unbalanced)
first, balance the equation
2Mg +O2-------> 2MgO
two magnesium atoms react with one diatomic oxygen molecule
there is a 1:1 ratio of magnesium to oxygen atoms
but we have 2 moles of magnesium atoms and 2X5 = 10 moles of oxygen atoms
the lesser magnesium LIMITS the amount of product we can make, so it is the LIMITING REAGENT.
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).
Altitude. Rainfall, a hot dry day, and a hurricane are all examples of weather, not climate.
<h3>
Answer:</h3>
42960 years
<h3>
Explanation:</h3>
<u>We are given;</u>
- Remaining mass of C-14 in a bone is 0.3125 g
- Original mass of C-14 on the bone is 80.0 g
- Half life of C-14 is 5370 years
We are required to determine the age of the bone;
- Remaining mass = Original mass × 0.5^n , where n is the number of half lives.
Therefore;
0.3125 g = 80.0 g × 0.5^n
3.90625 × 10^-3 = 0.5^n
- Introducing logarithm on both sides;
log 3.90625 × 10^-3 = n log 0.5
Solving for n
n = log 3.90625 × 10^-3 ÷ log 0.5
= 8
- Therefore, the number of half lives is 8
- But, 1 half life is 5370 years
- Therefore;
Age of the rock = 5370 years × 8
= 42960 years
Thus, the bone is 42960 years old