Answer:
13.5 g
Explanation:
This question is solved easily if we remember that the number of moles is obtained by dividing the mass into the atomic weight or molar mass depending if we are referring to elements or molecules.
Therefore, the mass of aluminum in the reaction will the 0.050 mol Al times the atomic weight of aluminum.
number of moles = n = mass of Al / Atomic Weight Al
⇒ mass Al = n x Atomic Weight Al = 0.050 mol x 27 g mol⁻¹
= 13.5 g
We have three significant figures in 0.050 and therefore we should have three significant figures in our answer.
Answer:
A fluid is a medium that has a defined mass and volume, but no fixed shape, at a constant temperature and pressure. This may include gases, liquids, plasmas, and to some extent plastic solids. A fluid can flow and deform, preventing it from carrying loads in a static equilibrium. A fluid is always compressible and internal frictional forces always occur due to the viscosity of the fluid.
Answer:
The solution is always homogeneous mixture and transparent through which the light can travel. The mixture of water and sugar is a solution because sugar is soluble in water and form homogeneous mixture while the sand can not dissolve in water and sand particles scatter the light.
Explanation:
Solution:
"The solution is always homogeneous mixture and transparent through which the light can travel"
The mixture of water and sugar is a solution because sugar is soluble in water and form homogeneous mixture. The solubility of sugar is high as compared to the sand in water because the negative and positive ends of sucrose easily dissolve into the polar solvent i.e, water
Suspension:
"Suspension is the heterogeneous mixture, in which the solute particles settle down but does not dissolve"
The mixture of water and sand is suspension. The sand can not dissolve in water because it is mostly consist of quartz. The nonpolar covalent bonds of sand are too strong and cannot be break by water molecules.
<span>pH is a measure of how acidic/basic water is.</span>
I don’t think so if you passed that class and your still doing course recovery you should talk to your counselor
