Answer:
state of matter
Explanation:
so take water for example, water has a melting point and a boiling point right? So if it's below 0 degrees, then it's in its solid phase. If the temperature is above 0 degrees, then the water starts to melt into its liquid phase. Then when the temperature is above 100 degrees, water starts to boil and become its gas phase. This is the same for all substances. The only difference is different substances have different melting and boiling points so the numbers will be different depending on your substance. hope this helped!
Answer:
No, ΔE does not always equal zero because it refers to the systems internal energy, which is affected by heat and work
Explanation:
According to the first law of thermodynamics, energy is neither created nor destroyed. This implies that the total energy of a system is always a constant.
So, according to the first law of thermodynamics we have that ΔE = q + w. This means that the value of ΔE depends on q (heat) and w(work). Hence ΔE is not always zero since it depends on the respective values of q and w.
Explanation:
here is the answer to your question
Answer:
Electron pair geometry- trigonal planar
There is one lone pair around the boron atom
The geometry of BH2 is bent
Explanation:
The valence shell electron pair repulsion theory offers a frame work for determining the shape of molecules based on the number of electron pairs of the valence shell of the central atom in the molecule.
In BH2-, the central atom is boron. There is a lone pair on boron. Owing to the lone pair on boron, the molecular geometry of BH2 is bent.
Answer:
3 Cu²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Cu₃(PO₄)₂(s)
Explanation:
Let's consider the molecular equation between aqueous copper(II) chloride and aqueous sodium phosphate.
3 CuCl₂(aq) + 2 Na₃PO₄(aq) ⇒ 6 NaCl(aq) + Cu₃(PO₄)₂(s)
The complete ionic equation includes all the ions and insoluble species.
3 Cu²⁺(aq) + 6 Cl⁻(aq) + 6 Na⁺(aq) + 2 PO₄³⁻(aq) ⇒ 6 Na⁺(aq) + 6 Cl⁻(aq) + Cu₃(PO₄)₂(s)
The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and insoluble species.
3 Cu²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Cu₃(PO₄)₂(s)
