Answer:
Element: Pure substance consisting of one type of atom. Compound: Pure substance consisting of two or more different atoms. Mixture: Two or more different substances not chemically combined.
Explanation:
PLEASE ADD ME AS BRAINLIEST
The thermochemical reaction for combustion of methane is the balanced chemical equation for the combustion reaction with the enthalpy change and states of the reactants included.
Combustion is when an organic compound reacts with O₂ to yield CO₂ and H₂O.
thermochemical reaction for combustion of methane is as follows;
CH₄(g) + 2O₂(g) --> CO₂(g) + 2H₂O(g) ΔH = -891 kJ/mol
For your first question, that equation only works if your situation is occurring at a constant temperature. Your original question is such a situation - everything occurs at 298.15 K. Therefore, you can use this value in the equation to calculate work.
For your second question, Charles' Law describes how the volume of gas changes as you heat or cool it, PROVIDED PRESSURE AND MOLES OF GAS REMAIN CONSTANT THE WHOLE TIME. In your original question above, temperature stays constant while volume changes. However, what they don't tell you is that this necessarily requires a change in either pressure or moles of gas. Because the question works with the same sample the of gas the whole time (i.e. moles are constant), it is pressure that is changing (and this change will occur according to Boyle's Law, since temperature and moles are held constant).
Hope that clarifies things!
Answer:
Boron Carbonate; B₂(CO₃)₃
Explanation:
For names, carbonide does not exist; that rules out the first option. Carbide refers to just a carbon atom, not carbon and oxygen as in the polyatomic ion carbonate. That rids us of the third option. We are left with boron carbonate with the formula BCO or boron carbonate with the formula B₂(CO₃)₃.
Recall the carbonate polyatomic ion's formula: CO₃²⁻
Thus BCO cannot be the formula.
Option 4 is your answer, Boron Carbonate; B₂(CO₃)₃.
To further check your answer, observe the oxidation states of boron and the polyatomic ion carbonate. Boron can exist in oxidation states of either 2+ or 3+, and carbonate is only 2-; in this formula, boron will exhibit a 3+ state to balance out with carbonate.
2x3+ = 6+; 3x2- = 6-
6+ + 6- = 0; balanced