Answer: The mowed lawn is the one from where the grasses are removed by using the machines or tools.
Explanation:
The mowed lawn is expected to have low number of species as the grasses may be few or scanty thus can support the population of few species like insects, mice, birds, and small number of grazing animals. On the other hand the weedy field can be hub of insects, reptiles like snakes, small mammals, and large mammals. Large weed field can provide food, and habitat to the large number of species. This will support the increase in biodiversity as compared to the mowed lawn.
Balanced equation is
HBr + NaOH ----> NaBr + H2O
Using molar masses
80.912 g HBr reacts with 39.997 g of Naoh to give 18.007 g water
so 1 gram of NaOH reacts with 2.023 g of HBR
and 5.7 reacts with 11.531 g HBr so we have excess HBr in this reaction
Mass of water produced = (5.7 * 18.007 / 39.997 = 2.6 g to 2 sig figs
Answer:
2nd option is NOT a postulate of kinetic theory of gases
Explanation:
actually particles move in straight random motion and create bombardment with the molecules and container
Answer:
Explanation:
Group one elements are alkali metals. All alkali metal have one valance electron. They loses their one valance electron and from cation with charge of +1.
Charges on group one.
Hydrogen = +1
Lithium = +1
Sodium = +1
Potassium = +1
Rubidium = +1
Cesium = +1
Francium = +1
Group two elements are alkaline earth metals. All alkaline earth metal have two valance electron. They loses their two valance electron and from cation with charge of +2.
Charges on group two.
Beryllium = +2
Magnesium = +2
Calcium = +2
Strontium = +2
Barium= +2
Radium = +2
Group 13 elements are boron family. All elements have three valance electrons. They loses their three valance electron and from cation with charge of +3.
Charges on group 13.
Boron = +3
Aluminium = +3
Gallium = +3
Indium = +3
Thallium= +3
Group 13 elements are also shows +1 charge by losing one valance electron.
Answer:
The balanced equation for this reaction is C2H2 + 502 + 4H2O + 3C02. What volume of carbon dioxide is produced when 2.8 L of oxygen are consumed? 25Explanation:
