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Blababa [14]
3 years ago
12

What is the net ionic charge of an oxygen ion?

Mathematics
2 answers:
bixtya [17]3 years ago
5 0

Answer:

One oxygen ion has a net ionic charge of -2

Step-by-step explanation:

One Oxygen ion has a total of 10 electrons (-10 charge), but only 8 protons (+8 charge). This calculates to a net ionic charge of -2. Hope this helps! :)

Paraphin [41]3 years ago
3 0

Answer: u need to go to mr.google sir/mam

Step-by-step explanation:

ur mom

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WILL<br><br><br> GIVE <br><br><br> BRAINLIST<br><br><br><br> ...<br> ..<br> MATH
Hatshy [7]

Answer:

according to meh it's -213

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8 0
2 years ago
Read 2 more answers
A radiator contains 10 quarts of fluid, 30% of which is antifreeze. How much fluid should be drained and replaced with pure anti
babymother [125]

Answer:

1 3/7 quarts should be drained off and replaced with pure antifreeze.

1 3/7 ≈ 1.4286

Current amount of antifreeze in quarts is -

30/ 100 × 10 = 3

40% ---> 4 quarts

Let the amount drained of and replaced with antifreeze be x-

The amount left after draining off is 10 − x.

The amount of antifreeze  is 30/ 100 (10−x).

30/100(10-x)+x=4

3-3/10x+x=4

3+x(1-3/10)=4

x=1*10/7=1 3/7 quarts

check;

10- 1 3/7 = 8 4/7

=(30/100*8 4/7)+1 3/7

=(3/10 * 60/7) + 10/7

=3*6/7 + 10/7

=28/7

=4

4 liters of pure antifreeze is mixed into 10 quarts.

8 0
3 years ago
ASAP PLEASE!!!
mario62 [17]
It should be D Vietnam I believe and by the way you have this under mathematics. 
8 0
3 years ago
Find the value of h
slamgirl [31]

Answer:

h = 48

because 32 + 3h + 4 = 180

32 + 4 = 36

36 + 3h - 36 = 180 - 36

3h = 144

3h/3 = 144/3

h = 48

7 0
3 years ago
Use the data table provided to calculated the values requested below. Provide all answers to three decimal places.
babymother [125]

Answer:

1) 0.700

2) 0.730

3) 0.030

4) 0.959

Step-by-step explanation:

1) proportion of support for the ban with at least one child =

     \frac{no of support atleast 1 child}{Total no of atleast 1 child\\}

   = \frac{1739}{2485}

   = 0.700

2)  proportion of support for the ban with no child =

    = \frac{no of support with no child}{Total of no child}

    = \frac{3089}{4231}

    = 0.730

3) Difference in proportion of supporters for the ban between those with atleast one child and those with no child

    =  0.700 - 0.730

    = -0.03

4) Relative risk = \frac{proportion with atleast on child}{proportion with no child}

    = \frac{0.700}{0.730}  = 0.959

5 0
3 years ago
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