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3 years ago

7

How do you know how many electrons an element will gain or lose in forming an ion?

1 answer:

Alinara [238K]3 years ago

7 0

First the element has to be stable. One chooses the easiest way to make the element stable. for example : Al has 13 electrons and you want to make it stable, you consider the outermost electrons where in AL the are three so it is easier for Al to lose three than to gain five

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Now that Snape and Dumbledore has taught you the finer points of hydration calculations they have a slightly more challenging pr

Mrrafil [7]

Answer:

The value of an integer x in the hydrate is 10.

Explanation:

$Molarity=\frac{Moles}{Volume(L)}$

Molarity of the solution = 0.0366 M

Volume of the solution = 5.00 L

Moles of hydrated sodium carbonate = n

$0.0366 M=\frac{n}{5.00 L}$

$n=0.0366 M\times 5 mol=0.183 mol$

Mass of hydrated sodium carbonate = n= 52.2 g

Molar mass of hydrated sodium carbonate = 106 g/mol+x18 g/mol

$n=\frac{\text{mass of Compound}}{\text{molar mass of compound}}$

$0.183 mol=\frac{52.2 g}{106 g/mol+x\times 18 g/mol}$

$106 g/mol+x\times 18 g/mol=\frac{52.2 g}{0.183 mol}$

Solving for x, we get:

x = 9.95 ≈ 10

The value of an integer x in the hydrate is 10.

6 0

3 years ago

Commercial concentrated aqueous ammonia is 28% nh3 by mass and has a density of 0.90 g/ml. what is the concentration of ammonia

patriot [66]

Answer:- 14.9 M

Solution:- Given commercial sample of ammonia is 28% by mass. Let's say we have 100 grams of the sample. Then mass of ammonia would be 28 grams.

Density of the solution is given as 0.90 grams per mL.

From the mass and density we could calculate the volume of the solution as:

$100g(\frac{1mL}{0.90g})$

= 111 mL

Let's convert the volume from mL to L as molarity is moles of solute per liter of solution.

$111mL(\frac{1L}{1000mL})$

= 0.111 L

Now, we convert grams of ammonia to moles on dividing the grams by molar mass. Molar mass of ammonia is 17 gram per mole.

$28g(\frac{1mole}{17g})$

= 1.65 mole

To calculate the molarity we divide the moles of ammonia by the liters of solution:

$molarity=\frac{1.65mole}{0.111L}$

= 14.9 M

So, the molarity of the given commercial sample of ammonia is 14.9 M.

5 0

3 years ago

Elements bond when they share or transfer their outer shell electrons which are called valence electrons

rusak2 [61]

Answer:

True

Explanation:

3 0

2 years ago

Find the number of kilograms in 15.40 moles of C5H10O5

OLga [1]

C5H10O5 would weight a total of 2.312 kilograms.

5 0

2 years ago

Can u pls help me with this question

lorasvet [3.4K]

Answer:

b i im pretty sure or a

Explanation:

f y bib0f84f69g85

8 0

2 years ago