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ratelena [41]
3 years ago
12

Iron (ii) hydroxide + sodium nitrate. Name the precipitate formed in reaction

Chemistry
1 answer:
olya-2409 [2.1K]3 years ago
4 0
Green colored precipitate of iron (||) hydroxide
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How do i balance C6H12O6+O2 > H2O+CO2
xenn [34]

Answer:

C6H12O6+6O2

Explanation:

c=6

H=12

O=6+2=8

C=1*6=6

H=2*6=12

O=2+1=3-->O=6(2)+6=18

6+?=18

?=12

5 0
3 years ago
An atom that ______ electrons is called a positive ion. A. has 0 B. has 8 C. loses D. gains
Kitty [74]

Answer:

Gains

Explanation:

It gets more electrons

8 0
3 years ago
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In the dark waters just below the photic zone, because of cellular respiration, the concentration of dissolved __________ is hig
Dafna1 [17]
In the dark waters just below the photic zone, because of cellular respiration, the concentration of dissolved carbon dioxide is higher relative to dissolved oxygen. 
In cellular respiration living organisms use oxygen and release carbon dioxide, because there is little or no light there is no photosynthesis and oxygen is little produced.
7 0
3 years ago
A student measures the length of a 1.00 m standard bar. He obtains measurements of 0.99 m, 1.01 m, and 1.00 m. Which describes h
scoray [572]
The correct answer is (D. both precise and accurate.
8 0
3 years ago
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Citrate synthase catalyzes the reaction: ????x????????o????c???????????????????? + ????c????????y???? − ????o???? → c???????????
Nady [450]

The given question is incomplete. The complete question is as follows.

Citrate synthase catalyzes the reaction

Oxaloacetate + acetyl-CoA \rightarrow citrate + HS-CoA

The standard free energy change for the reaction is -31.5 kJ*mol^-1

( a) Calculate the equilibrium constant for this reaction a 37degrees C

Explanation:

(a).  It is known that , relation between change in free energy (\Delta G) of a reaction and equilibrium constant (K) is as follows.

             \Delta G = -RT \times ln K  

where,  T = temperature in Kelvin

The given data is as follows.

         T = 310 K,       \Delta G = -31.5 kJ /mol = -31500 J/mol  (as 1 kJ = 1000 J)

Now, putting the given values into the above formula as follows.

     ln K = \frac{-(\Delta G)}{RT}

            = \frac{31500}{8.314 \times 310}

      ln K = 12.22

         K = antilog (12.22)

           = 2.1 \times 10^{5}

Therefore, we can conclude that value of equilibrium constant for the given reaction is 2.1 \times 10^{5}.

6 0
3 years ago
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