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olga55 [171]
3 years ago
7

All scientists try to base their conclusions on

Chemistry
2 answers:
lana66690 [7]3 years ago
6 0
<span>All scientists try to base their conclusions on the experiments they have conducted and they have repeated them to be sure of it. </span>
Inessa05 [86]3 years ago
3 0
The information drawn from completing lab test and proven data from tests and hypothesis 
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In order to perform a chemical reaction, 225 mL of 0.500 M lead (II) nitrate is required. How much 5.00 M stock solution is requ
vladimir1956 [14]

Answer:

what

Explanation:

what ````

4 0
3 years ago
Which element decreases its oxidation number in this reaction? bicl2 + na2so4 → 2nacl + biso4?
kifflom [539]
The balanced reaction is as follows;
BiCl₂ + Na₂SO₄ --> 2NaCl + BiSO₄
this is a double displacement reaction 
the oxidation number of Bi is +2 in both BiCl₂ and BiSO₄
oxidation number of Cl is -1 in both BiCl₂ and NaCl 
oxidation number of Na is +1 in both Na₂SO₄ and NaCl
oxidation numbers of elements in SO₄²⁻ remains the same in both compounds.Therefore the oxidation state in any of the elements in the reaction doesn't change. Neither of the elements show an increase or decrease in the oxidation numbers .
Answer for this question is no element decreases its oxidation number.

5 0
3 years ago
Read 2 more answers
Which molecule or reaction supplies the energy for polymerization of nucleotides in the process of transcription?
Sindrei [870]

Answer to this is "the phosphate bonds that are present in the nucleotide triphosphate serves as the substrates" in the process of transcription.

Transcription is the process in which DNA is transformed into RNA with the help of an enzyme known as RNA-polymerase. RNA and DNA are the nucleic acids which are formed by linking the nucleotides together.


7 0
3 years ago
True or false. (Pls help)
grigory [225]
I believe they’re both true.
5 0
2 years ago
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What are the products obtained in the electrolysis of molten nai?
yanalaym [24]
Answer is: sodium (Na) and iodine (I₂).

<span> First ionic bonds in this salt are separeted because of heat: 
</span>NaI(l) → Na⁺(l) + I⁻(l).

Reaction of reduction at cathode(-): Na⁺(l) + e⁻ → Na(l) /×2.

2Na⁺(l) + 2e⁻ → 2Na(l).

Reaction of oxidation at anode(+): 2I⁻(l) → I₂(l) + 2e⁻.

The anode is positive and the cathode is negative.


4 0
3 years ago
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