So if you put a holo ball of paper in the water it sinks and as it sinks it gets smaller cuse pressure so it cuseus things to shrink cuse of pressure
The answer is B as isotopes are different versions of the same chemical element containing the same amount of protons and electrons but different amounts of neutrons.
Answer:
The percent by mass of 3.55 g NaCl dissolved in 88 g water is 3.88%
Explanation:
When a solute dissolves in a solvent, the mass of the resulting solution is a sum of the mass of the solute and the solvent.
A percentage is a way of expressing a quantity as a fraction of 100. In this case, the percentage by mass of a solution is the number of grams of solute per 100 grams of solution and can be represented mathematically as:
In this way it allows to precisely establish the concentration of solutions and express them in terms of percentages.
In this case:
- mass of solute: 3.55 g
- mass of solution: 3.55 g + 88 g= 91.55 g
Replacing:
Percent by mass= 3.88%
<u><em>The percent by mass of 3.55 g NaCl dissolved in 88 g water is 3.88%</em></u>
Answer:
See explanation
Explanation:
The reactivity of metals has a lot to do with their position in the electrochemical series. However, it is also known that metallic character decreases across the period. This implies that as we move from left to right along the periodic table. Sodium, magnesium, aluminum and silicon continues to decrease in metallic character. As a matter of fact, silicon is a metalloid and not a pure metal.
Sodium reacts with cold water to give a vigorous reaction,magnesium and aluminium reacts with steam at red heat.
Silicon does not react with water, even as steam, under normal conditions.
Answer:
Limiting reactant: O2
grams NO2 produced = 230.276 g NO2
grams of NO unused = 26.67 gNO
Explanation:
2NO + O2 --> 2NO2
Step 1: Determine the molar ratio NO:O2
molar ratio NO:O2 = 5.895: 2.503 = 2.35
stoichiometric molar ratio NO:O2 = 2:1
So, O2 is the limiting reactant.
Step2: Determine the grams of NO2:
?g NO2 = moles O2 x (2moles NO2/1 mol O2) x (MM NO2/ 1 mol NO2) = 2.503 x 2 x 46 = 230.276 g NO2
Step 3: Determine the amount of excess reagent unreacted
moles excess NO reacted = moles O2 x (2 moles NO/1 mol O2) = 2.503 x 2 = 5.006 moles NO reacted
moles NO unreacted = total moles NO - moles NO reacted = 5.895-5.006 =0.889 moles NO unreacted
mass NO unreacted = moles NO unreacted x MM NO = 0.889 x 30 =26.67 g NO unreacted
