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faltersainse [42]
2 years ago
5

Exercise 2: The words below are chopped into pieces. Find

Chemistry
1 answer:
Elanso [62]2 years ago
7 0

Answer:

Weather, environment, ecosystem, species, rain forests

Explanation:

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An acid with molar mass 84.48 g/mol is titrated with 0.650 M KOH. What volume of KOH solution is needed to titrate 1.70 grams of
RSB [31]

Answer:

V=0.0310L=3.10mL

Explanation:

Hello.

In this case, since the acid is monoprotic and the KOH has one hydroxyl ion only, we can see that at the equivalence point the moles of both of them are the same:

n_{acid}=n_{KOH}

Thus, since we are given 1.70 g of the acid, we compute the moles of acid that were titrated:

n_{acid}=1.70g*\frac{1mol}{84.48g}=0.0201mol

Which equal the moles of KOH. In such a way, since the molarity is defined as moles over liters (M=n/V), the liters are moles over molarity (V=n/M), thus, the resulting volume is:

V=\frac{0.0201mol}{0.650mol/L}\\\\V=0.0310L=3.10mL

Best regards!

7 0
2 years ago
Given the balanced equation below, calculate the moles of aluminum that are needed to react completely with 28.7 moles of FeO. Y
Alexxx [7]

Answer:

43.05 moles of Al needed to react with 28.7 moles of FeO.

Explanation:

Given data:

Moles of FeO = 28.7 mol

Moles of Al needed to react with FeO = ?

Solution:

Chemical equation:

2Al + 3FeO → 3Fe + Al₂O₃

Now we will compare the moles of Al with FeO.

                            FeO        :           Al

                             2            :            3

                          28.7          :         3/2×28.7 = 43.05 mol

Thus 43.05 moles of Al needed to react with 28.7 moles of FeO.

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Concave Lenses

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