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7nadin3 [17]
3 years ago
10

For a school event 1/6 of the athletic field is reversed for the fifth -grade classes the reserved part of the field is divided

equally Among the 4 fifth grade classes in the school what fraction of the whole athletic field is reserved for each fifth class
Chemistry
1 answer:
SVETLANKA909090 [29]3 years ago
3 0

Answer:

\frac{1}{24}

Explanation:

Given:

For a school event, 1/6 of the athletic field is reserved for the fifth -grade classes and the reserved part of the field is divided equally among the 4 fifth grade classes in the school.

To find: fraction of the whole athletic field reserved for each fifth class

Solution:

Fraction of the whole athletic field reserved for four fifth classes = \frac{1}{6}

So, fraction of the whole athletic field reserved for each fifth class = \frac{1}{4}(\frac{1}{6})=\frac{1}{24}

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Explanation:

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with 6 H2, 4O2 is excess.

H2O molecules formed = 6

7 0
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Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho
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Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0
3 years ago
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3 0
3 years ago
Which liquids would cause the largest decrease in mass of a potato stick
riadik2000 [5.3K]

Answer:

distilled water I guess !

Just a guess though

If you find this useful, please mark my answer as the brainliest.

Explanation:

If you find this useful, please mark my answer as the brainliest.

distilled water

Explanation:

Some students investigated osmosis in raw potato sticks. The students measured the mass of three potato sticks using an electronic balance. The students left each potato stick in one of the three different liquids for 5 hours:i. distilled water. ii. dilute sodium chloride solution. iii. concentrated sodium chloride solution. After 5 hours they measured the mass again and calculated the change in mass. 1. Predict which of the liquids would cause the largest decrease in mass of a potato stick. 2. After the experiment, the students noticed that the potato stick with the lowest mass was soft and floppy. Explain why the potato stick had become soft and floppy. 3. The students followed the same experimental procedure with boiled potato sticks and found no overall change in mass in any of the solutions. Suggest why the mass of the boiled potato sticks remained the same.

3 0
3 years ago
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