Answer:
See explanation
Explanation:
A. This is a neutralization reaction.
Molecular equation;
HBr(aq) + CsOH(aq) ---------> CsBr(aq) + H20(l)
Complete ionic equation;
H^+(aq) + Br^-(aq) + Cs^(aq) + OH^-(aq) --------> Cs^+(aq) + Br^- + H20(l)
Net ionic equation;
H^+(aq) + OH^-(aq) --------> H20(l)
B. This is a gas forming reaction;
H2SO4(aq) + Na2CO3(aq) ------->Na2SO4(aq) + H2O(l) + CO2(g)
Complete ionic equation;
2H^+(aq) + SO4^-(aq) + 2Na^+(aq) + CO3^2-(aq) ------->2Na^+(aq) + SO4^-(aq) + H2O(l) + CO2(g)
Net ionic equation;
2H^+(aq) + CO3^2-(aq) -------> + H2O(l) + CO2(g)
C. This a precipitation reaction
Molecular equation;
CdCl2(aq) + Na2S(aq) ------->CdS(s) + 2NaCl(aq)
Complete ionic equation;
Cd^2+(aq) + 2Cl^-(aq) + 2Na^+(aq) + S^2-(aq) ---------> CdS(s) + 2Na^+(aq) + 2Cl^-(aq)
Net ionic equation;
Cd^2+(aq) + S^2-(aq) ---------> CdS(s)
Answer:
A) The mass would be the same.
Explanation:
Since there is no loss of any particle to vapor during the phase change process from solid to liquid, the mass of the before and after the process will remain the same.
- In this way, the law of conservation of mass is obeyed.
- Mass is the amount of matter contained in a substance.
- Since there is no room for escape or matter loss, the mass will remain the same.
Air is mainly composed of N2 (78%), O2 (21%) and other trace gases. Now, the total pressure of air is the sum of the partial pressures of the constituent gases. The partial pressure of each gas, for example say O2, can be expressed as:
p(O2) = mole fraction of O2 * P(total, air) ----(1)
Thus, the partial pressure is directly proportional to the total pressure. If we consider a sealed container then, as the temperature of air increases so will its pressure. Based on equation (1) an increase in the pressure of air should also increase the partial pressure of oxygen.
Answer:
pH = 6.82
Explanation:
To solve this problem we can use the<em> Henderson-Hasselbach equation</em>:
- pH = pKa + log
![\frac{[NaOCl]}{[HOCl]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNaOCl%5D%7D%7B%5BHOCl%5D%7D)
We're given all the required data to <u>calculate the original pH of the buffer before 0.341 mol of HCl are added</u>:
- pKa = -log(Ka) = -log(2.9x10⁻⁸) = 7.54
- [HOCl] = [NaOCl] = 0.500 mol / 0.125 L = 4 M
- pH = 7.54 + log
By adding HCl, w<em>e simultaneously </em><u><em>increase the number of HOCl</em></u><em> and </em><u><em>decrease NaOCl</em></u>:
- pH = 7.54 + log
![\frac{[NaOCl-HCl]}{[HOCl+HCl]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNaOCl-HCl%5D%7D%7B%5BHOCl%2BHCl%5D%7D)
- pH = 7.54 + log
There are 453.592 grams in a pound.
