Answer:
Density
Explanation:
Density is defined as the mass per unit volume. It is the ratio between the mass and the volume of a substance. It does not matter how large or small a sample of matter is, the same substance will always have the same density, because of this. The ratio between the mass and volume remains the same.
I believe the correct answer from the choices listed above is option D. The action that leads to crystal formation in minerals is that atoms or molecules form repeating patterns. Minerals are known to have a crystalline structure in which they exhibit short range and long range patterns.
Answer:
H^+(aq) + OH^-(aq) -------> H2O(l)
Explanation:
We must first write the molecular reaction equation as follows;
HNO3(aq) + NaOH(aq) ------>NaNO3(aq) + H2O(l)
The complete ionic equation is;
H^+(aq) + NO3^-(aq) + Na^+(aq) + OH^-(aq) -------> Na^+(aq) + NO3^-(aq) + H2O(l)
The net ionic equation therefore is;
H^+(aq) + OH^-(aq) -------> H2O(l)
Answer: A monopoly is the absence of competition in the market.
Explanation:
In such circumstances, the market creates a monopoly of one producer who takes huge capital and dictates prices. An example of a monopoly on the market is the existence of only one company that makes up the entire economic branch. In such circumstances, the monopolist can increase the product's price without losing the entire sale, i.e., operating successfully. In that situation, the monopolist remains the only one on the market, and the competition has no access to the market.
Answer:
B) ) –1615.1 kJ mol^–1
Explanation:
since
SiO2(s) + 4 HF(aq) → SiF4(g) + 2 H2O(l) ∆Hºrxn = 4.6 kJ mol–1
the enhalpy of reaction will be
∆Hºrxn = ∑νp*∆Hºfp - ∑νr*∆Hºfr
where ∆Hºrxn= enthalpy of reaction , ∆Hºfp= standard enthalpy of formation of products , ∆Hºfr = standard enthalpy of formation of reactants , νp=stoichiometric coffficient of products, νr=stoichiometric coffficient of reactants
therefore
∆Hºrxn = ∑νp*∆Hºfp - ∑νr*∆Hºfr
4.6 kJ/mol = [1*∆HºfX + 2*(–285.8 kJ/mol)] - [1*(–910.9kJ/mol) + 4*(–320.1 kJ/mol)]
4.6 kJ/mol =∆HºfX -571.6 kJ/mol + 2191.3 kJ/mol
∆HºfX = 4.6 kJ/mol + 571.6 kJ/mol - 2191.3 kJ/mol = -1615.1 kJ/mol
therefore ∆HºfX (unknown standard enthalpy of formation = standard enthalpy of formation of SiF4(g) ) = -1615.1 kJ/mol
