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V125BC [204]
3 years ago
6

Calcule la normalidad de una solución

Chemistry
1 answer:
tensa zangetsu [6.8K]3 years ago
5 0

Answer:

si por favor

Explanation:

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Complete this equation for the dissociation of na2co3(aq). omit water from the equation because it is understood to be present.
Savatey [412]
Net overall dissociation:
Na2CO3 ---> 2Na(-) + CO3(2-)

*The ion charge is in parenthesis
6 0
3 years ago
Read 2 more answers
A
Artemon [7]

Answer:

9.8ms^{-2}

Explanation:

using the equation :

acceleration = \frac{final  velocity - initial velocity}{timetaken} \\                      a  =\frac{v-u}{t}

where v is 98m/s

u is 0m/s

t is 10seconds

a=\frac{98m/s- 0m/s}{10} \\a=\frac{98}{10}\\ a=9.8m/s^{2}

3 0
3 years ago
Could someone explain how they got this answer, explain step by step plz
gulaghasi [49]

Answer:

6.018 amu

Explanation:

Let 6–Li be isotope A.

Let 7–Li be isotope B.

Let the abundance of 6–Li be A%

Let the abundance of 7–Li be B%

The following data were obtained from the question:

Atomic mass of isotope A (6–Li) =.?

Atomic mass of isotope B (7–Li ) = 7.015 amu.

Abundance of 7–Li (B%) = 92.58%

Abundance of 6–Li (A%) = 100 – B% = 100 – 92.58 = 7.42%

Atomic mass of Lithium = 6.941amu

The atomic mass of isotope A (6–Li) can be obtained as follow:

Atomic mass = [(Mass of A x A%)/100] + [(Mass of B x B%)/100]

6.941 = [(mass of A x 7.42)/100] + [(7.015x92.58)/100]

6.941 = [(mass of A x 7.42)/100] + 6.494487

(mass of A x 7.42)/100 = 6.941 – 6.494487

(mass of A x 7.42)/100 = 0.446513

Mass of A x 7.42 = 100 x 0.446513

Mass of A x 7.42 = 44.6513

Divide both side by 7.42

Mass of A = 44.6513 / 7.42

Mass of A = 6.018 amu

Therefore, the mass of 6–Li is 6.018 amu

7 0
3 years ago
Network forensics might deal with what? Retrieving photos from a PC Analyzing text messages Investigating a data breach Defragme
balandron [24]

Answer:

Investigating a data breach

Explanation:

A data breach usually involves data exfiltration over a computer network. the other options involve data being stored on a device locally which isn't volatile data like text messages, photos or rearranging data in defragmentation all of which does not require a network.

8 0
3 years ago
Hurry!! Please
Vanyuwa [196]

c

Explanation:

pay attention in class next time bye hope it helps THE ANSWER IS C bye

7 0
2 years ago
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