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ratelena [41]
3 years ago
13

In sodium chloride, which atom loses an electron?

Chemistry
2 answers:
s344n2d4d5 [400]3 years ago
5 0
B. sodium
explanation:
sodium is in front (NaCl), and the first element always loses an electron
Fittoniya [83]3 years ago
3 0

Answer:

Option B. Sodium

Explanation:

In the formation of NaCl, sodium loses and electron while chlorine receives the electron as illustrated below:

Na —> Na+ + e- ....... (1)

Cl + e- —> Cl- ......... (2)

Now, combining equation 1 and 2 we have:

Na + Cl + e- —> Na+ + Cl- + e-

Cancelling the electron from both side, we obtained:

Na + Cl —> Na+Cl-

From the above illustration, we see clearly that sodium loses electron

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Which of the following is true for the endothermic reaction Ba(OH)2(aq) + NH4NO3(aq) --> Ba(NO3)2(aq) + 2NH3(aq) + 2H2O(l)?
Alchen [17]
If it's an endothermic reaction, then that means heat is being added to the system therefore H>0. Entropy is disorder, and since there are more moles on the products side, entropy is increasing therefore S>0 as well. 
6 0
3 years ago
The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151
vitfil [10]

Answer:-  The natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for ^1^5^1_E_u and ^1^5^3_E_u as 150.919860 and 152.921243 amu, respectively.  Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of ^1^5^1_E_u as n then the abundance of ^1^5^3_E_u would be 1-n .

Let's plug in the values in the formula:

151.964=150.919860(n)+152.921243(1-n)

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

151.964-152.921243=150.919860n-152.921243n

-0.957243=-2.001383n

negative sign is on both sides so it is canceled:

0.957243=2.001383n

n=\frac{0.957243}{2.001383}

n=0.478

The abundance of ^1^5^1_E_u is 0.478 which is 47.8%.  

The abundance of ^1^5^3_E_u is = 1-0.478

= 0.522 which is 52.2%

Hence, the natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .


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