Answer:
The Empirical Formula.
Explanation:
From the empirical formula and using the weight (in g) of a given substance, we can come up with the molecular formula which is the actual weight of a substance. Sometimes, we find that the empircal formula is the molecular formula.
Answer:
Like other alkali metals, rubidium metal reacts violently with water. As with potassium (which is slightly less reactive) and caesium (which is slightly more reactive), this reaction is usually vigorous enough to ignite the hydrogen gas it produces.
Explanation:
hope it helps
The first step is to find the number of moles of OH⁻ that reacted with the HCl. To do this multiply 2.00L by 1.50M to get 3 moles of Ca(OH)₂. Then you multiply 3 by 2 (there are 2 moles of OH⁻ per every 1 mole of Ca(OH)₂) to get 6 moles of OH⁻. That means that you needed 6 moles of HCl since 1 mole of HCl contains 1 mole of H⁺ and equal amounts H⁺ and OH⁻ reacted with each other. To find the molarity of the HCl solution you need to divide 6mol by 1L to get 6M. Tat means that the concentration of the acid was 6M.
I hope this helps. Let me know if anything was unclear.
Answer:
a. in supernovae and star collisions
Explanation:
The periodical table contains some heavier elements, which are formed as neutron stars pairs hit eachother and erupt cataclysmically.
The star emitts very large quantities of energy and neutrons during supernova, which allow for the production of heavier elements than iron, such as uranium and gold. All these elements are ejected into space during the supernova explosion.
Answer:
The answer to your question is pH = 0.686
Explanation:
Data
Acetic acid = 20 ml
Final volume = 1.7 L
pH = ?
density = 1.05 g/ml
Process
1.- Calculate the mass of acetic acid
density = mass / volume
mass = density x volume
mass = 1.05 x 20
mass = 21 g
2.- Calculate the moles of acetic acid (CH₃COOH)
Molar mass = (12 x 2) + (16 x 2) + (4 x 1)
= 24 + 32 + 4
= 60 g
60 g of acetic acid ---------------- 1 mol
21 g of acetic acid ----------------- x
x = (21 x 1) / 60
x = 0.35 moles
3.- Calculate the concentration
Molarity = 0.35 moles / 1.70 L
= 0.206
4.- Calculate the pH
pH = -log[0.206]
pH = 0.686
