V1 = 2.00 L
<span>T1 = 25 + 273 = 298 K </span>
<span>V2 = 6.00 L </span>
<span>T2 = ? </span>
<span>Assuming the pressure is to remain constant, then </span>
<span>V1/T1 = V2/T2 </span>
<span>T2 = T1V2/V1 = (298)(6)/(2) = 894 deg K</span>
Answer:
3.2 × 10⁻⁸
Explanation:
Let's consider the solution of magnesium carbonate.
MgCO₃ ⇄ Mg²⁺(aq) + CO₃²⁻(aq)
We can relate the molar solubility (S) with the solubility product (Ksp) using an ICE chart.
MgCO₃ ⇄ Mg²⁺(aq) + CO₃²⁻(aq)
I 0 0
C +S +S
E S S
The Ksp is:
Ksp = [Mg²⁺] × [CO₃²⁻] = S × S = S² = (1.8 × 10⁻⁴)² = 3.2 × 10⁻⁸
Answer:
C. Its oxidation number increases.
Explanation:
- <em><u>Oxidation is defined as the loss of electrons by an atom while reduction is the gain of electrons by an atom</u></em>.
- Atoms of elements have an oxidation number of Zero in their elemental state.
- When an atom looses electrons it undergoes oxidation and its oxidation number increases.
- For example, <em><u>an atom of sodium (Na) at its elemental state has an oxidation number of 0. When the sodium atom looses an electrons it becomes a cation, Na+, with an oxidation number of +1 , the loss of electron shows an increase in oxidation number from 0 to +1.</u></em>
Explanation:
The electrons in the outermost shell of an atom are the valence electrons. These electrons are the most loosely held in an atom.
The energy required to remove these electrons are not as great as those of the inner shell electrons.
- Atoms of some elements show no tendency to combine with other atoms because they have completely filled outer energy levels.
- Their outermost shell, valence shell and the electron numbers are complete for them. These are the noble gases.
- other atoms share or exchange their valence electrons in order to have a stable configuration.
- The valence electrons are involve in inter-atomic bond formation and interactions
- Also the distribution of these electrons determines inter-molecular interactions between molecules.
Learn more:
valence electrons in metalloids brainly.com/question/3023499
#learnwithBrainly
, THR CC14 formed in the first step is used as the reactant used in the second step.if 5.00 mol of CH4 reacts, what is the total amount of HCI producded. assume that C12 an HR in the presentin excess