Neutrons don’t carry an electrical charge, meaning that adding or subtracting them from the nucleus will not change the electrical charge of the nucleus of an atom. But, adding/removing neutrons changes the mass of the nucleus. This is how isotopes are formed.

5 0

3 years ago

All atoms of a given element are identical was proven wrong because of the existence or

vagabundo [1.1K]

Answer:

they are all similar

Explanation:

6 0

3 years ago

Quinine, an antimalarial drug, is 8.63% nitrogen. There are two nitrogen atoms per molecule. What is the molecular weight of qui

Furkat [3]

Answer:

324.18 g/mol

Explanation:

Let the molecular mass of the antimalarial drug, Quinine is x g/mol

According to question,

Nitrogen present in the drug is 8.63% of x

So, mass of nitrogen = $\frac {8.63}{100}\times x$

Also, according to the question,

2 atoms are present in 1 molecule of the drug.

Mass of nitrogen = 14.01 amu = 14.01 g/mol (grams for 1 mole)

So, mass of nitrogen = 14.01×2 = 28.02

These 2 must be equal so,

$\frac {8.63}{100}\times x=28.02$

solving for x, we get:

x = 324.18 g/mol

6 0

4 years ago

Based on the sign of E cell, classify these reactions as spontaneous or non spontaneous as written.? assume standard conditions.

sammy [17]

A electrochemical reaction is said to be spontaneous, if $E^{0} cell is positive.$

Answer 1:
Consider reaction: Ni^2+ (aq) + S^2- (aq) ----> + Ni (s) + S (s)

The cell representation of above reaction is given by;
 $S^{2-}/S // Ni^{2+}/Ni$

Hence, $E^{0}cell = E^{0} Ni^{2+/Ni} - E^{0} S/S^{2-}$
we know that, ${E^{0} Ni^{2+}/Ni = -0.25 v$
and ${E^{0} S/ S^{2-} = -0.47 v$

Therefore, $E^{0} cell$ = - 0.25 - (-0.47) = 0.22 v

Since, $E^{0} cell$ is positive, hence cell reaction is spontaneous
.....................................................................................................................

Answer 2:
Consider reaction: Pb^2+ (aq) +H2 (g) ----> Pb (s) +2H^+ (aq)

The cell representation of above reaction is given by;
 $H_{2} / H^{+} // Pb^{2+} /Pb$

Hence, $E^{0}cell = E^{0} Pb/Pb^{2+} - E^{0} H_{2}/H^{+}$
we know that, ${E^{0} Pb^{2+}/Pb = -0.126 v$
and ${E^{0} H_{2}/ H^{+} = -0 v$

Therefore, $E^{0} cell$ = - 0.126 - 0 = -0.126 v

Since, $E^{0} cell$ is negative, hence cell reaction is non-spontaneous.

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Answer 3:
Consider reaction: 2Ag^+ (aq) + Cr(s) ---> 2 Ag (s) +Cr^2+ (aq)

The cell representation of above reaction is given by;
 $Cr/Cr^{2+} // Ag^{+}/Ag$

Hence, $E^{0}cell = E^{0} Ag^{+}/Ag - E^{0} Cr/Cr^{2+}$
we know that, ${E^{0} Ag^{+}/Ag = -0.22 v$
and ${E^{0} Cr/ Cr^{2+} = -0.913 v$

Therefore, $E^{0} cell$ = - 0.22 - (-0.913) = 0.693 v

Since, $E^{0} cell$ is positive, hence cell reaction is spontaneous

8 0

4 years ago

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