6g of hydrogen gas is my answer. I'm sorry if I'm wrong.
<span>283.89 g/mol is the molar mass of tetraphosphorus decoxide</span>
If the dehydration reaction of an alcohol is successful. The changes would be seen in the IR spectrum for the product compared to the starting material are as,
- The O-H and C-O band is disappear from stating material
- The addition of a C-C double bond band in the product.
In dehydration reaction of alcohol ( O-H and C-O bond ) contain , the water molecule (
) is release from the reactant and C-C double bond is form which is known as alkene in the product .
The reactant and product have different structure. To determine the structure of the compound IR spectroscopy is used. In IR spectrum the peak corresponds to 3400-3600 cm is missing in the product of dehydration reaction of an alcohol. It means O-H band is disappear from stating material.
learn about IR SPECTRUM
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Answer:
Aluminum metal
Explanation:
In order to properly answer this or a similar question, we need to know some basic rules about galvanic cells and standard reduction potentials.
First of all, your strategy would be to find a trusted source or the table of standard reduction potentials. You would then need to find the half-equations for aluminum and gold reduction:


Since we have a galvanic cell, the overall reaction is spontaneous. A spontaneous reaction indicates that the overall cell potential should be positive.
Since one half-equation should be an oxidation reaction (oxidation is loss of electrons) and one should be a reduction reaction (reduction is gain of electrons), one of these should be reversed.
Thinking simply, if the overall cell potential would be obtained by adding the two potentials, in order to acquite a positive number in the sum of potentials, we may only reverse the half-equation of aluminum (this would change the sign of E to positive):
Notice that the overall cell potential upon summing is:

Meaning we obey the law of galvanic cells.
Since oxidation is loss of electrons, notice that the loss of electrons takes place in the half-equation of aluminum: solid aluminum electrode loses 3 electrons to become aluminum cation.
Answer:
320 g
Step-by-step explanation:
The half-life of Co-63 (5.3 yr) is the time it takes for half of it to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows:
No. of Fraction Mass
half-lives t/yr Remaining Remaining/g
0 0 1
1 5.3 ½
2 10.6 ¼
3 15.9 ⅛ 40.0
4 21.2 ¹/₁₆
We see that 40.0 g remain after three half-lives.
This is one-eighth of the original mass.
The mass of the original sample was 8 × 40 g = 320 g