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Kisachek [45]
3 years ago
6

Friction, conduction, and induction are all methods through which _______ can be transferred.

Chemistry
2 answers:
Ugo [173]3 years ago
8 0

Friction, conduction, and induction are all methods through which energy can be transferred.

Hope that helps.

GREYUIT [131]3 years ago
4 0

Answer:

It is charge.

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Explain why the product at the negative electrode is not always a metal.
Elena-2011 [213]

Answer:

Whether you get the metal or hydrogen during electrolysis depends on the position of the metal in the reactivity series: the metal will be produced if it is less reactive than hydrogen. hydrogen will be produced if the metal is more reactive than hydrogen.

3 0
3 years ago
How does a molecule differ from an atom?
Eddi Din [679]

Answer:

According to my research A molecule is two or more atoms held together by covalent bonds. An atom is the smallest part of an element. ... A sodium atom has one outer electron, and a carbon atom has four outer electrons.

Explanation:

5 0
3 years ago
If you knew the number of valence electrons in a nonmetal atom how would you determine the valence of the element. (Ignore hydro
SIZIF [17.4K]

Answer:

The possible valances can be determined by electron configuration and electron negativity

Good Luck even though this was asked 2 weeks ago

Explanation:

All atoms strive for stability. The optima electron configuration is the electron configuration of the VIII A family or inert gases.

Look at the electron configuration of the nonmetal and how many more electrons the nonmetal needs to achieve the stable electron configuration of the inert gases. Non metals tend to be negative in nature and gain electrons. ( They are oxidizing agents)

For example Florine atomic number 9 needs one more electron to reach a valance number of 8 electrons to equal Neon atomic number 10. Hence Flowrine has a valance of -1

Oxygen atomic number 8 needs two more electrons to reach a valance number of 8 electrons to equal Neon atomic number 10. Hence Oxygen has a valance charge of -2.

Non metals with a low electron negativity will lose electrons when reacting with another non metal that has a higher electron negativity. When the non metal forms an ion it is necessary to look at the electron structure to determine how many electrons the element can lose to gain stability.

For example Chlorine which is normally -1 like Florine when it combines with oxygen can be +1, +3, + 5 or +7. It can lose its one unpaired electron, or combinations of the unpaired electron and sets of the three pairs of electrons.

6 0
2 years ago
Complete the sentences about heme. Some terms will not be used. The prosthetic group of hemoglobin and myoglobin is . The organi
nevsk [136]

Explanation:

Haemoglobin consists of heme unit which is comprised of an <u>Fe^{2+}</u> and porphyrin ring. The ring has four pyrrole molecules which are linked to the iron ion. In oxyhaemoglobin, the iron has coordinates with four nitrogen atoms and one to the F8 histidine residue and the sixth one to the oxygen. In deoxyhaemoglobin, the ion is displaced out of the ring by 0.4 Å.

The prosthetic group of hemoglobin and myoglobin is - <u>Heme</u>

The organic ring component of heme is - <u>Porphyrin</u>

Under normal conditions, the central atom of heme is - <u>Fe^{2+}</u>

In <u>deoxyhemoglobin</u> , the central iron atom is displaced 0.4 Å out of the plane of the porphyrin ring system.

The central atom has <u>six</u> bonds: <u>four</u> to nitrogen atoms in the porphyrin, one to a <u>histidine</u> residue, and one to oxygen.

7 0
3 years ago
A solution of NaOH is titrated with H2SO4. It is found that 20.05 mL of 0.3564 M H2SO4 solution is equivalent to 43.42 mL of NaO
Darya [45]

Answer : The concentration of NaOH is, 0.336 M

Explanation:

To calculate the concentration of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=0.3564M\\V_1=20.05mL\\n_2=1\\M_2=?\\V_2=43.42mL

Putting values in above equation, we get:

2\times 0.3564M\times 20.05mL=1\times M_2\times 43.42mL

M_2=0.336M

Thus, the concentration of NaOH is, 0.336 M

3 0
3 years ago
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