For the first question, salt is soluble while sand is insoluble or not dissolvable in water. The salt should have vanished or melted, but the sand stayed noticeable or visible, making a dark brown solution probably with some sand particles caught on the walls of the container when the boiling water was put in to the mixture of salt and sand. The solubility of a chemical can be disturbed by temperature, and in the case of salt in water, the hot temperature of the boiling water enhanced the salt's capability to melt in it.
For the second question, the melted or dissolved salt should have easily made its way through the filter paper and into the second container, while the undissolved and muddy sand particles is caught on the filter paper. The size of the pores of the filter paper didn’t change. On the contrary, the size of the salt became smaller because it has been dissolved which is also the reason why it was able to go through the filter paper, while the size of the sand may have doubled or even tripled which made it harder to pass through.
Answer:
A; or carnivores that feed on producers
Explanation:
Answer:
3.8 x 10⁵
Explanation:
For the equilibrium : 3NO(g) ⇌ N2O(g) + NO2(g), the equilibrium constant in the terms of the concentrations of the gases in mol/L is
Kc = (NO) (N2O)/ (NO) ³ where (NO), (N2O) , (NO2) are the concentrations of the gases in mol/L . So
K= (x mol/ 1 L) (x mol/1L) / (7.5 x 10⁻⁶ mol /1 L) ³
x = mol of NO and NO2 at equilibrium
we have that
K = x²/ 7.5 x 10⁻⁶ = 1.9 x 10¹⁶
x = √ (7.5 x 10⁻⁶ x 1.9 x 10¹⁶) = 3.8 x 10⁵
∴ (N2O) = 3.8 x 10⁵
As the warm water holds more salt then cold water the fingers cool and produce crystals of salt that soon rains down to the floor of the ocean
Answer:
8.5gm O2 produced
Explanation:
When heated, KClO3 decomposes into KCl and O2. 2KClO3⟶2KCl+3O2 If this reaction produced 13.2 g KCl, how many grams of O2 were produced?
for every 2 moles of KCl produced, 3 moles of O2 are produced
Mol weight of KCl =39+35.5=74.5gm
13.2 gm KCl = 13.2/74.5 = 0.177 moles
this will make (3/2) X 0.177 = 0.2655 moles of O2
O2 mol wt is 32 0.2655 X32 = 8.5gm O2 produced
