The correct answer is 1. Lose electrons and become positive ions.
I hope my answer was beneficial to you! c:
When the reaction equation is:
CaSO3(s) → CaO(s) + SO2(g)
we can see that the molar ratio between CaSO3 & SO2 is 1:1 so, we need to find first the moles SO2.
to get the moles of SO2 we are going to use the ideal gas equation:
PV = nRT
when P is the pressure = 1.1 atm
and V is the volume = 14.5 L
n is the moles' number (which we need to calculate)
R ideal gas constant = 0.0821
and T is the temperature in Kelvin = 12.5 + 273 = 285.5 K
so, by substitution:
1.1 * 14.5 L = n * 0.0821 * 285.5
∴ n = 1.1 * 14.5 / (0.0821*285.5)
= 0.68 moles SO2
∴ moles CaSO3 = 0.68 moles
so we can easily get the mass of CaSO3:
when mass = moles * molar mass
and we know that the molar mass of CaSO3= 40 + 32 + 16 * 3 = 120 g/mol
∴ mass = 0.68 moles* 120 g/mol = 81.6 g
Answer:
Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.
Explanation:
Volume of NaOH = 1.7 ml = 0.0017 L
Molarity of NaOH = 0.0811 M
Moles of NaOH = n
n = 0.0001378 mol

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.
Then 0.0001378 mol of NaOH will neutralize:
of sulfuric acid.
Concentration of sulfuric acid in the acid rain sample: x

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.
<u>Answer:</u> The enthalpy of the reaction for the production of
is coming out to be -74.9 kJ
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28reactant%29%7D%5D)
For the given chemical reaction:

The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CH_4(g))})]-[(1\times \Delta H^o_f_{(C(s))})+(2\times \Delta H^o_f_{(H_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CH_4%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C%28s%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2%28g%29%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times (-74.9))]-[1\times 0)+(2\times 0)]\\\\\Delta H^o_{rxn}=-74.9kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-74.9%29%29%5D-%5B1%5Ctimes%200%29%2B%282%5Ctimes%200%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-74.9kJ)
Hence, the enthalpy of the reaction for the production of
is coming out to be -74.9 kJ
Answer:
The half-life varies depending on the isotope.
Half-lives range from fractions of a second to billions of years.
The half-life of a particular isotope is constant.