Answer is: <span>No, because she did not stop adding base once the color changed.
</span>The endpoint<span> is the point at which the indicator changes colour in a colourimetric </span>titration and that is point when titration must stop or results are going to be wrong, because t<span>he </span><span>equivalence point of titration is not measured right.</span>
C. crop rotation
Explanation:
Crop rotation is a farming practice that is commonly used to maintain a healthy soil.
This practices helps to conserve and appropriate a soil for the best use.
During crop rotation, different crop species are planted during a growing season. The patterns are designed to cultivate both crops that replenish and deplete the nutrients in soil.
Through this the fertility of the soil is sustained all year round.
This helps to avoid the use of pesticides on crops.
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Answer: condensaytion
Explanation: its reverse vaporation instead of water turning into gas gas turns into water
Answer:
Wowww this is cool did you draw this?
Explanation:
brainlest po
We have to first write a balanced equation.
so2 + o2 -> so3
this is not balanced though. we have 3 oxygen on right and 4 on left
2so2 + o2 -> 2so3
now it is same on both sides. we have to figure out which is limiting reagent with the given amounts of reagents. we do this by comparing the ratio between them in terms of moles. we see that so2 has a coefficient of 2 and o2 has none which implies 1 and so3 has 2. this means that for every 2 moles of so2 reacting with 1 mole of o2, we get 2 moles of so3.
lets convert the given values to moles. to do this we know that molecular weight is measured in grams per mole. we are given grams and need to cancel out the grams to get moles. so the molecular weight:
so2 =32.1 + 2 * 16 = 64.1 g/mol
o2 = 2 * 16 = 32 g/mol
so3 = 32.1 + 3 * 16 = 80.1 g/mol
now to convert 90 g of 2so2 under ideal conditions.
90g / 64.1g/mol = 1.404 moles
convert this amount of moles of so2 to moles of o2. we have 2 moles of so2 to 1 of o2
1.404moles so2 / 2 moles so2 * 1 mole o2= 0.702 moles o2
so we see under ideal conditions that 90g of so2 would react with .702g of o2. lets see how many we actually have with 100g of o2
100g / 32g/mol =3.16 mol.
so we have a lot more o2 than needed. we are looking for how much is left in grams. we have to figure out how much was used. to do this convert our ideal moles of o2 into grams.
.702 moles o2 * 32g/mol = 22.5g o2
so what we startrd with (100g) minus what we needed (22.5g) is what we have left
100 - 22.5 = 77.5g o2
