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Studentka2010 [4]
3 years ago
15

Which statement is true of rectangles Y and Z? Rectangle Y has a length of 6 and width of 4. Rectangle Z has a length of 4 and w

idth of 3. They are congruent because their corresponding angles are congruent. They are similar because their corresponding angles are congruent. They are similar because their corresponding side lengths are proportional. They are not similar because their corresponding side lengths are not proportional.
Chemistry
2 answers:
zmey [24]3 years ago
8 0

Answer:

b

Explanation:

Free_Kalibri [48]3 years ago
4 0

Answer: Rectangle Z has a length of 4 and width of 3. They are congruent because their corresponding angles are congruent. They are similar because their corresponding angles are congruent. They are ... Rectangle Y has a length of 6 and width of 4.

Explanation: HOPE IT HELP YOU

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Answer:

A

Explanation:

The number of protons and neutrons of an element is the same. the electrons are the only thing that can differ.  The atomic number equal the protons and neutrons.

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Which of the following is true if the net force on an object is zero? (1 pt) a. The object must be at rest. b. The object’s spee
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Explanation:

the object must be at rest

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Julio earns $300 a week working a summer job. The paycheck he deposits in the bank is for $250. Which of these accounts for the
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theirs a $50 tax with the income hes making

Explanation:

5 0
3 years ago
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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
Sophie [7]

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
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