The Avogadro number is 6 times 10^{23}. What | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The atomic mass of $_6C^{14}$ is $14 \, g/mol$.The number of moles in $14 \, g$ of $_6C^{14}$ is $n = \frac{14 \, g}{14 \, g/mol} = 1 \, mol$.The

Vedclass · Accepted Answer

$36 	imes 10^{23}, 48 	imes 10^{23}, 36 	imes 10^{23}$

Vedclass · Answer

(A) The atomic mass of $_6C^{14}$ is $14 \, g/mol$.The number of moles in $14 \, g$ of $_6C^{14}$ is $n = \frac{14 \, g}{14 \, g/mol} = 1 \, mol$.The number of atoms in $1 \, mol$ is equal to the Avogadro number,$N_A = 6 	imes 10^{23}$.In one atom of $_6C^{14}$:Protons $(Z)$ = $6$Neutrons $(N = A - Z)$ = $14 - 6 = 8$Electrons = $6$ (since it is a neutral atom).Total number of protons = $6 	imes (6 	imes 10^{23}) = 36 	imes 10^{23}$.Total number of neutrons = $8 	imes (6 	imes 10^{23}) = 48 	imes 10^{23}$.Total number of electrons = $6 	imes (6 	imes 10^{23}) = 36 	imes 10^{23}$.

List-$I$	List-$II$
$(A)$ Strong nuclear force	$(i)$ $10^{-2}$
$(B)$ Weak nuclear force	(ii) $1$
$(C)$ Electromagnetic force	(iii) $10^{-39}$
$(D)$ Gravitational force	(iv) $10^{-13}$

The Avogadro number is $6 \times 10^{23}$. What will be the number of protons,neutrons,and electrons in $14 \, g$ of $_6C^{14}$?

Similar Questions

If the nucleus ${}_{13}^{27}Al$ has a nuclear radius of about $3.6 \, fm$,then ${}_{52}^{125}Te$ would have its radius approximately as .......$fm$.

Match the following fundamental forces of nature with their relative strength:
List-$I$ List-$II$
$(A)$ Strong nuclear force $(i)$ $10^{-2}$
$(B)$ Weak nuclear force (ii) $1$
$(C)$ Electromagnetic force (iii) $10^{-39}$
$(D)$ Gravitational force (iv) $10^{-13}$

Which of the following pairs is an isobar?

Isotopes are atoms having

The volume of a nucleus is directly proportional to

Vedclass Test Series

Exam Paper Generator

Online Exam Module