Answer: 2.7 centimeters long.
Explanation: count from starting at the beginning of the number on the ruler "2" and counted up until the end of the blue object. Counted 7 lines, so, 2.7 is the correct answer.
Fluorine (F) would be least likely to form a cation out of potassium, fluorine, chlorine, and nitrogen.
- A cation is a positively charged atom (or molecule) that has lost electrons (or electrons).
- The tendency for electropositive elements to lose electrons and produce cations is greater. On the left side of the periodic table, these are often metals.
- Going down a group increases electropositivity, or the propensity to lose electrons and generate cations. and decreases across a period. In the given examples:
- Potassium, K is an alkali metal and will lose electrons readily to form a cation.
- Nitrogen (N), Fluorine (F), and chlorine (Cl) are all nonmetals that prefer to accept electrons and form anions instead. F is the most electronegative i.e. it will gain electrons and form F- rather than F+.
To learn more about cation visit:
brainly.com/question/28710898
#SPJ4
The reactant is Mercury (II) Oxide while the products are Mercury and Oxygen separately.
This is because the reactants are typically always on the left side of the yields symbol. In this decomposition reaction, it would still be the same as at the end of the reaction, there were to products produced: Mercury and Oxygen.
Products tend to always be on the right side of the yields symbol, they're what comes out of a reaction no matter what type.
Hope this helps!
D. The particles are tightly packed together
The reaction corresponds to the combustion of propane (C3H8). The balanced reaction is:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
The reaction enthalpy is given as:
ΔHrxn = ∑nΔH°f(products) - ∑ nΔH°f(reactants)
= [3ΔH°f(CO2(g)) + 4ΔH°f(H2O(g)] - [1ΔH°f(C3H8(g)) + 5ΔH°f(O2(g)]
= [3(-393.5) + 4(-241.8)] - [-103.9 + 5(0)] = -2043.8 kJ
The enthalpy for the combustion of propane is -2044 kJ
