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BARSIC [14]
3 years ago
14

Which part of John Dalton’s theory was disproved by J.J. Thomson’s cathode-ray tube experiments?(1 point)

Chemistry
1 answer:
andrezito [222]3 years ago
4 0
Atoms are indivisible.
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When water decomposes into oxygen and hydrogen, the mass
Maurinko [17]
<span>When water decomposes into oxygen and hydrogen, the mass "Remains Constant" as according to Law of Conservation of mass, mass can neither be created not destroyed,.

In short, Your Answer would be Option A

Hope this helps!</span>
8 0
3 years ago
What is a covalent bond?
Sphinxa [80]
It is the electron sharing.
electronegative element + electronegative element

exemple :

O₂ , H₂

hope this helps!
3 0
3 years ago
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
3 years ago
How do the equilibrium concentrations of the reactants compare to the equilibrium concentrations of the product?
Scorpion4ik [409]

Answer: It depends equilibrium constant K

Explanation:   You need to to have reaction formula.

If K >> 1 then concentrations of products are much bigger than

concentrations of reactants. If K < < 1, concentration of products is small.

6 0
3 years ago
What is the mass in g of 7.009 x 10^21 molecules of Pl3?
ycow [4]
I have completed the problem

4 0
2 years ago
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