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natulia [17]
3 years ago
11

Substances that contain a transition metal can be oxidizing. The greater the number of oxidation of this metal, the greater the

oxidizing character of the substance. In an industrial process in which the use of an oxidizing agent is necessary, only five substances are available: CaO, FeO, Cu₂O, Cr₂O₃ and KMnO₄.
The substance that should be used in this process, as it has a greater oxidizing character, is:


a) Cr₂O₃


b) Cu₂O


c) CaO


d) FeO


e) KMnO₄
Chemistry
1 answer:
Ne4ueva [31]3 years ago
7 0
B I think that it but I don’t know
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Consider the partial equation. Br2 + NaF --> Which best completes the chemical equation?
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Answer is (1) - no reaction.


<em>Explanation; </em>


Some of you may think this reaction as a single replacement reaction which gives NaBr + F₂ as products.


But, according to the reactivity of the halogens, reactivity decreases from up to bottom of the group. F is placed above Br. Hence, F is more reactive than Br. Hence, Br can't replace F.




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3 years ago
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Why did the French set up trading posts in North America?
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Answer: Their main aim was to provide goods, specifically furs and sugar exportation.

Explanation:

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Given an initial cyclopropane concentration of 0.00560 m, calculate the concentration of cyclopropane that remains after 1.50 ho
maria [59]

<span>We can solve this problem by assuming that the decay of cyclopropane follows a 1st order rate of reaction. So that the equation for decay follows the expression:</span>

A = Ao e^(- k t) 

Where,

A = amount remaining at time t = unknown (what to solve for) <span>
Ao = amount at time zero = 0.00560 M </span><span>
<span>k = rate constant
t = time = 1.50 hours or 5400 s </span></span>

The rate constant should be given in the problem which I think you forgot to include. For the sake of calculation, I will assume a rate constant which I found in other sources:

k = 5.29× 10^–4 s–1                     (plug in the correct k value)

<span>Plugging in the values in the 1st equation:</span>

A = 0.00560 M * e^(-5.29 × 10^–4 s–1 * 5400 s )

A = 3.218 <span>× 10^–4 M           (simplify as necessary)</span>

8 0
3 years ago
Which of the following is a type of electromagnetic radiation?
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6 0
3 years ago
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A compound containing sodium, chlorine, and oxygen is 25.42% sodium by mass. A 3.25 g sample gives 4.33×1022 atoms of oxygen. Wh
Tcecarenko [31]
step  one 
calculate  the  %  of  oxygen
from  avogadro  constant
1moles =  6.02  x  10  ^23  atoms
what  about    4.33  x10^22  atoms
= ( 4.33  x  10^ 22 x 1 mole )  /  6.02  10^23=   0.0719 moles
mass=  0.0719  x16=  1.1504   g
% composition   is therefore= ( 1.1504/3.25)  x100 = 35.40%
 step  two
calculate the  %  composition  of  chrorine
100-  (25.42  +  35.40)=39.18%

step  3
calculate the  moles   of  each  element
that   is  
Na  =  25.42  /23=1.1052  moles
Cl=  39.18  /35.5=1.1037moles
O=  35.40/16=  2.2125   moles
step  4
find  the  mole  ratio  by  dividing  each  mole  by  1.1037  moles
that  is
Na  =  1.1052/1.1037=1.001
Cl= 1.1037/1.1037=  1
0=2.2125 = 2
therefore  the  empirical  formula= NaClO2
8 0
4 years ago
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