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alisha [4.7K]
2 years ago
12

11. Do the blue dots on the graph represent an object that

Chemistry
1 answer:
dolphi86 [110]2 years ago
5 0
I think it’s C
I’m so sorry if it’s wrong!
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Write the equilibrium constant expression, Kc, for the following reaction: Please enter the compounds in the order given in the
xeze [42]

Answer:

See the answer below, please.

Explanation:

The equilibrium constant is defined as the relationship between products and reagents, each one elevated to their stoichiometric coefficients, in that of the given equation, the Kc is:

Kc= (NH4)^1/ (NH3)^1 x (HI)^1

NH4= products

NH3 and HI = reagents

7 0
3 years ago
What is very fine sediment called
Alecsey [184]
Loess, which can be carried by wind over long distances.
4 0
2 years ago
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How can a material get a charge?
svet-max [94.6K]

Explanation:

By losing or gaining electrons from its outermost orbit

8 0
2 years ago
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Fe(s) + O2 (g) → Fe3+ O2− What most likely happens during this reaction?
Musya8 [376]

Answer : The most likely happens during this reaction is, Oxidation-reduction

Explanation :

The balanced reaction will be,

4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)

In this reaction, neutral iron loses 3 electrons and oxidizes in (+3) state, Fe^{3+} and neutral oxygen gains 2 electrons and reduces in (-2) state,O^{2-}

When iron react with oxygen gas to give iron oxide. This process is known as iron rusting. During the reaction, oxidation-reduction process occurs.

Oxidation : It is a type of chemical reaction in which a substance loses its electrons. Or we can say that in oxidation, the oxidation number increases.

Reduction : It is a type of chemical reaction in which a substance gains its electrons. Or we can say that in reduction, the oxidation number decreases.

5 0
3 years ago
How many grams of glucose are needed to prepare 400. g of a 2.00% (m/m) glucose solution g?
Dominik [7]
The grams   of glucose  are  needed  to  prepare  400g  of  a 2.00%(m/m)  glucose  solution  g  is  calculated  as  below

=% m/m =mass  of the solute/mass  of  the  solution  x100

let mass of   solute  be represented  by  y
mass  of solution = 400 g
 % (m/m) = 2% = 2/100

 grams  of  glucose  is  therefore =2/100 =  y/400
by cross  multiplication

100y = 800
divide   both side  by  100

y= 8.0 grams



5 0
3 years ago
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