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Zepler [3.9K]
3 years ago
5

Iron combines with 5 g of Copper (1) Nitrate to form 6.01 g of Iron (1) Nitrate and 0.4 g of copper

Chemistry
1 answer:
Ivanshal [37]3 years ago
7 0
First, you would add the products: 0.4 + 6.01 = 6.41 g
Then, to get the mass of Iron required, you would do 6.41 - 5.0 = 1.41
This is because in the conservation of mass principle it states,
mass of reactants = mass of products.
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How many electrons are transferred during the formation of one molecule of al2o3?
lukranit [14]
Answer:
            Six electrons are transferred in the formation of Al₂O₃.

Explanation:
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                                2 Al  +  3/2 O₂    →    Al₂O₃

Oxidation number of Al on left hand side is zero, while than on right hand side in Al₂O₃ is +3. Means it has lost 3 electrons per one atom and six electrons per two atoms. Also, the oxidation number of O at left hand side in O₂ is zero, while that in Al₂O₃ it is -2 per atom and -6 per 3 atoms. 

So, two Al atoms have lost 6 electrons and 3 O atoms have gained six electrons.
4 0
3 years ago
The 10x SDS gel electrophoresis buffer contains 250mM Tris HCl, 1.92M Glycine, and 1% (w/v) SDS. Buffers are always used at 1x c
olchik [2.2K]

Answer:

25 mM Tris HCl and 0.1% w/v SDS

Explanation:

A <em>10X solution</em> is ten times more concentrated than a <em>1X solution</em>. The stock solution is generally more concentrated (10X) and for its use, a dilution is required. Thus, to prepare a buffer 1X from a 10X buffer, you have to perform a dilution in a factor of 10 (1 volume of 10X solution is taken and mixed with 9 volumes of water). In consequence, all the concentrations of the components are diluted 10 times. To calculate the final concentration of each component in the 1X solution, we simply divide the concentration into 10:

(250 mM Tris HCl)/10 = 25 mM Tris HCl

(1.92 M glycine)/10 = 0.192 M glycine

(1% w/v SDS)/10 = 0.1% w/v SDS

Therefore the final concentrations of Tris and SDS are 25 mM and 0.1% w/v, respectively.

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