Answer:
Modern science is typically divided into three major branches that consist of the natural sciences (biology, chemistry, physics, astronomy and Earth science), which study nature in the broadest sense; the social sciences (e.g. psychology, sociology, economics, history) which study people and societies; and the formal ...
Alkaline earth metals are metals of group two. They are divalent metals and they have a highly negative reduction potential hence the metals are mostly extracted by electrolysis.
They are highly reactive metals. They react with water but do so less readily than alkali earth metals.
Owing to their high reactivity, they are seldom found free in nature. They always occur in combined state with other highly reactive nonmetals.
Answer:
38.541 × 10¹⁹ formula units
Explanation:
Given data:
Mass of chromium sulfate = 0.25 g
Formula units in 0.25 g = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ formula units of water
Number of moles of chromium sulfate = Mass / molar mass
Number of moles of chromium sulfate = 0.25 g/ 392.16 g/ mol
Number of moles of chromium sulfate = 6.4 × 10⁻⁴ moles
Number of formula units:
1 mole = 6.022 × 10²³ formula units
6.4 × 10⁻⁴ moles × 6.022 × 10²³ formula units / 1 mol
38.541 × 10¹⁹ formula units
False. An increase in temperature is an exothermic reaction. However, when a temperature decreases this is known as an endothermic reactionz
Answer:
0.42 M
Explanation:
The reaction that takes place is:
- Cu(CH₃COO)₂ + Na₂CrO₄ → Cu(CrO₄) + 2Na(CH₃COO)
First we <u>calculate the moles of Na₂CrO₄</u>, using the <em>given volume and concentration</em>:
(200 mL = 0.200L)
- 0.70 M * 0.200 L = 0.14 moles Na₂CrO₄
Now we <u>calculate the moles of Cu(CH₃COO)₂</u>, using its <em>molar mass</em>:
- 40.8 g ÷ 181.63 g/mol = 0.224 mol Cu(CH₃COO)₂
Because the molar ratio of Cu(CH₃COO)₂ and Na₂CrO₄ is 1:1, we can directly <u>substract the reacting moles of Na₂CrO₄ from the added moles of Cu(CH₃COO)₂</u>:
- 0.224 mol - 0.14 mol = 0.085 mol
Finally we <u>calculate the resulting molarity</u> of Cu⁺², from the <em>excess </em>cations remaining:
- 0.085 mol / 0.200 L = 0.42 M