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3 years ago

Given the formulas for the ionic compounds, draw the correct ratio of cations to anions (a) BaSO4, (b) CaF2, (c) Mg3N2, (d) K2O.

Chemistry

1 answer:

Karolina [17]3 years ago

3 0

Answer:

A. It formed by barium(Ba+2) ion and sulfate ( SO42- )

B. It is formed by calcium ion (Ca+2) and two fluoride ions (2F-)

C. It is formed by magnesium ion (Mg+2) and nitride ion (N3-)

D. It is formed by two potassium ions (2 K+) and oxide ion(O2-)

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Suppose that a certain fortunate person has a net worth of $79.0 billion ($7.90×1010). If her stock has a good year and gains $3

mario62 [17]

A net worth: $79.0 billion.
Value of stock : $3.20 billion.
New net worth:
$79.0 + $3.20 = $82.20 billion = $82.20 * 10^9 = $8.20 * 10^10

7 0

4 years ago

How many moles of zinc are in 4.70 kg of zinc?

faust18 [17]

First we convert kg to g
1kg =1000g
4.70=x(gram)
4.70×1000÷1=4700g
now:
number of moles=mass÷molarmass
number of moles=4700÷65.4
number of moles=71.86moles.

6 0

3 years ago

Read 2 more answers

Easy points!

podryga [215]

The soda an air is most likely bubbling up and is going to explode.

3 0

3 years ago

What mass of HCL is consumed by the reaction of 5. 0 of magnesium?

nignag [31]

I’m assuming you mean 5g of magnesium :)

4 0

3 years ago

Given that a chlorine-oxygen bond has an enthalpy of 243 kJ/mol , an oxygen-oxygen bond has an enthalpy of 498 kJ/mol , and the

Alecsey [184]

Explanation:

The chemical equation is as follows.

$\frac{1}{2}Cl_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow ClO(g)$

And, the given enthalpy is as follows.

$\frac{1}{2}Cl_{2}(g) + O_{2}(g) \rightarrow ClO_{2}(g)$ ; $\Delta H$ = 102.5 kJ

Cl-Cl = 243 kJ/mol, O=O = 498 kJ/mol

Since, the bond enthalpy of Cl-Cl is not given so at first, we will calculate the value of Cl-Cl as follows.

$\Delta H = \sum \text{bond broken in reactants} - \sum \text{bond broken in products}$

102.5 = $[(\frac{1}{2})x + 498] - [(2)(243)]$

102.5 = $(\frac{1}{2})x + 498 - 486$

102.5 - 12 = $\frac{x}{2}$

x = 181 kJ

Now, total bond enthalpy of per mole of ClO is calculated as follows.

$\Delta H = \sum E(\text{bond broken in reactants}) - \sum (\text{bond broken in products})$

x = $[(\frac{1}{2})181 + (\frac{1}{2})498] - 243$

= 339.5 - 243

= 96.5 kJ

Thus, we can conclude that the value for the enthalpy of formation per mole of ClO(g) is 96.5 kJ.

8 0

3 years ago