Answer:
2 electrons
8 electrons
8 electrons
1 electron (just count in the rings for these ones)
19 protons (it says "P=19" in the middle. The P stands for Protons)
K (it's from Latin medieval Latin "kalium")
39 (add proton and neutron amounts to get the approximate atomic mass--electrons are negligible)
Explanation:
Answer:
True
Explanation:
One of the strong proponent to support a hypothesis is its ability to be falsifiable. Also, a hypothesis must be testable. To truly test a hypothesis, one must be able to prove it wrong. If a hypothesis cannot be proven wrong, then it is a very good hypothesis and can be made a theory. A good hypothesis is one that cannot be falsifiable and can be easily tested in an experimental set up.
Answer:
<em>Please note that the question is incorrectly formatted. It should read:</em>
“How many moles of Na 2 CO 3 are in 10.0 mL of a 2.0 M solution?” (Note the subscripts. Without them, the question makes no sense.)
When a solution is called 2.0 M (molar), it has 2 moles of compound dissolved in 1 liter of water. So how many moles would there be in 10 ml?
(1L =) 1000 ml = 2 moles
10 ml = x moles.
Cross-multiply: x = 10*2/1000 = 0.02 moles.
(Or alternatively, 10 ml is 100 times less than 1 L so you also have 100 times less moles.)
Group names in the periodic table give clues about the metallic properties of the elements.
Metallic elements are found on the left side of the periodic table. A simple conception of metals describes them as a lattice of positive ions immersed in a sea of electrons.
Answer:
1. Mg (s) + 2Na+(aq) → 2Na(s) + Mg²⁺(aq)
2. 2K(s) + Cd²⁺(aq) → 2K⁺(aq) + Cd(s)
Explanation:
The net ionic equation of a reaction express only the chemical species that are involved in the reaction:
1. Mg (s) + Na2CrO4 (aq) → 2Na + MgCrO4(aq)
The ionic equation:
Mg (s) + 2Na+(aq) + CrO4²⁻ (aq) → 2Na + Mg²⁺ + CrO4²⁻(aq)
Subtracting the ions that don't change:
<h3>Mg (s) + 2Na+(aq) → 2Na + Mg²⁺</h3>
2. 2K(s) + Cd(NO3)2(aq) → 2KNO3(aq) + Cd(s)
The ionic equation:
2K(s) + Cd²⁺(aq) + 2NO3⁻(aq) → 2K⁺(aq) + 2NO3⁻(aq) + Cd(s)
Subtracting the ions that don't change:
<h3>2K(s) + Cd²⁺(aq) → 2K⁺(aq) + Cd(s)</h3>