It seems more and more there are fewer conservation organizations who speak for the forest, and more that speak for the timber industry. Witness several recent commentaries in Oregon papers that are by no means unique. I’ve seen similar themes from other conservation groups across the West in recent years.
Many conservation groups have uncritically adopted views that support more logging of our public lands based upon increasingly disputed ideas about forest health and fire ecology, as well as the age-old bias against natural processes like wildfire and beetles.
For instance, an article in the Portland Oregonian quotes Oregon Wild’s executive director Sean Stevens bemoaning the closure of a timber mill in John Day Oregon. Stevens said: “Loss of the 29-year-old Malheur Lumber Co. mill would be ‘a sad turn of events’” Surprisingly, Oregon Wild is readily supporting federal subsidies to promote more logging on the Malheur National Forest to sustain the mill.
Answer:
74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.
Explanation:
The balanced reaction is:
Na₂CO₃ + Ca(NO₃)₂ ⟶ CaCO₃ + 2 NaNO₃
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:
- Na₂CO₃: 1 mole
- Ca(NO₃)₂: 1 mole
- CaCO₃: 1 mole
- NaNO₃: 2 mole
Being the molar mass of the compounds:
- Na₂CO₃: 106 g/mole
- Ca(NO₃)₂: 164 g/mole
- CaCO₃: 100 g/mole
- NaNO₃: 85 g/mole
then by stoichiometry the following quantities of mass participate in the reaction:
- Na₂CO₃: 1 mole* 106 g/mole= 106 g
- Ca(NO₃)₂: 1 mole* 164 g/mole= 164 g
- CaCO₃: 1 mole* 100 g/mole= 100 g
- NaNO₃: 2 mole* 85 g/mole= 170 g
You can apply the following rule of three: if by stoichiometry 106 grams of Na₂CO₃ produce 100 grams of CaCO₃, 79.3 grams of Na₂CO₃ produce how much mass of CaCO₃?

mass of CaCO₃= 74.81 grams
<u><em>74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.</em></u>
The question is incomplete, complete question is:
When copper(I) sulfide is partially roasted in air (reaction with oxygen), copper(I) sulfite is formed first. subsequently, upon heating, the copper sulfite thermally decomposes to copper(I) oxide and sulfur dioxide. Write balanced chemical equations for these two reactions.
Answer:
The balanced chemical equations for these two reactions:


Explanation:
On partial roasting of copper sulfide in an air. The balanced chemical reaction is given as:

On further heating of copper(I) sulfite it get decomposes into copper oxide and sulfur dioxide. The balanced chemical reaction is given as:
