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2 years ago

What would the formula of an ionic compound made of Aluminum and Oxygen be?

Chemistry

1 answer:

yan [13]2 years ago

8 0

Answer:

Sodium Aluminate

Formula:-

$\quad{:}\longrightarrow\sf NaAlO_2$

<h3>Explore more:-</h3>

Melting points and boiling points of some ionic compounds

$\boxed{\begin{array}{|c|c|c|c|} \cline {1-4}\Large\sf {Ionic\:Compounds} & \sf \Large {Chemical\:Formula} & \Large \sf{Melting\:point_{(k)}} & \Large\sf {Boiling\:point_{(k)}} \\ &&& \\ \cline {1-4} \sf Sodium\:Chloride & \sf NaCl & 1074 & 1686 \\ &&& \\ \cline {1-4} \sf Lithium\:Chloride & \sf \sf LiCl & 887 & 1600 \\ &&& \\ \cline {1-4} \sf Calcium\:Chloride & \sf CaCl_2 & 1045 & 1900 \\ &&& \\ \cline {1-4} \sf Calcium\:Oxide & \sf CaO & 2850 & 3120 \\ &&& \\ \cline {1-4} \sf Magnesium\:Chloride & \sf MgCl_2 & 981 & 1685 \\ &&& \\ \cline {1-4} \end{array}}$

You might be interested in

What are the products of the reaction with their coefficients C2H18+O2?

lozanna [386]

Answer:

2C2H18 + 13O2 → 4CO2 + 18H2O (I balanced your equation)

Products - CO2 & H20

4 0

2 years ago

How would you prepare 500 ml of the following solutions : Sodium succinate buffer (0.1 mol/dm3 pH 5.64)

Nana76 [90]

Answer:

8.10g of sodium succinate must be added and 247mL of 0.1M HCl adding enough water until make 500mL

Explanation:

Succinic acid has a pKa₂ of 5.63

To solve this question we must find the amount of sodium succinate and 0.1M HCl that we have to add using H-H equation:

pH = pKa + log [A-] / [HA]

5.64 = 5.63 + log [Na₂Succ.] / [HSucc⁺]

0.01 = log [Na₂Succ.] / [HSucc⁺]

1.0233 = [Na₂Succ.] / [HSucc⁺] (1)

As:

0.1M = [Na₂Succ.] + [HSucc⁺] (2)

Replacing (2) in (1):

1.0233 = 0.1M - [HSucc⁺] / [HSucc⁺]

1.0233[HSucc⁺] = 0.1M - [HSucc⁺]

2.033[HSucc⁺] = 0.1M

[HSucc⁺] = 0.0494M

[Na₂Succ] = 0.0506M

Both [Na₂Succ⁺] and [HSucc⁺] ions comes from the same sodium succinate we have to find the moles of sodium succinate in 500mL of 0.1M. Then, based on the reaction:

Na₂Succ + HCl → HSucc⁺ + Cl⁻

The moles of HCl added = Moles HSucc⁺ we need:

Moles Na₂Succ:

0.500L * (0.1mol/L) = 0.0500 moles

Mass -Molar mass sodium succinate: 162.05g/mol-:

0.0500mol * (162.05g/mol) = 8.10g of sodium succinate must be added

Moles HCl:

0.0494M * 0.500L = 0.0247 moles HCl * (1L / 0.1mol) = 0.247L =

And 247mL of 0.1M HCl adding enough water until make 500mL

7 0

3 years ago

For the following reaction, 35.4 grams of zinc oxide are allowed to react with 6.96 grams of water . zinc oxide(s) + water(l) --

IRISSAK [1]

Answer:

$m_{Zn(OH)_2}=38.4g$

Explanation:

Hello!

In this case, for the undergoing chemical reaction:

$ZnO(s)+H_2O(l)\rightarrow Zn(OH)_2$

We evaluate the yielded moles of zinc hydroxide by each reactant as shown below:

$n_{Zn(OH)_2}^{by ZnO}=35.4gZnO*\frac{1molZnO}{81.38gZnO}*\frac{1molZn(OH)_2}{1molZnO} =0.435molZn(OH)_2\\\\n_{Zn(OH)_2}^{by H_2O}=6.96gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{1molZn(OH)_2}{1molH_2O} =0.386molZn(OH)_2$

In such a way, since the water yields a smaller amount of zinc hydroxide we conclude it is the limiting reactant so the maximum mass is computed below:

$m_{Zn(OH)_2}=0.386molZn(OH)_2*\frac{99.424 gZn(OH)_2}{1molZn(OH)_2} \\\\m_{Zn(OH)_2}=38.4g$

Because the water limits the yielded amount of zinc hydroxide.

Best regards!

5 0

2 years ago

How do the bonds differ in each type?

VashaNatasha [74]

Answer:

Bonds basically differs with each other due to sharing of electrons .

Explanation:

There are majorly three kinds of bonds

1. Ionic bonds which forms due to an element donate an electron to another element completely .

2. covalent bonds which forms with the mutual sharing of electrons b/w two atoms .

3. metallic bonds which forms b/w metals & they share electrons due to electron negativity difference b/w two atoms or elements

7 0

2 years ago

Read 2 more answers

At 300.0 K and 0.987 atm pressure, what will be the volume of 2.30 mol of Ne?

Nadya [2.5K]

Answer:

V = 57.39 L

Explanation:

Given that,

Temperature, T = 300 K

Pressure, P = 0.987 atm

No. of moles of Ne, n = 2.30 mol

We need to find the volume of Ne. We know that, the ideal gas law is as follows :

PV = nRT

Where

P is pressure and R is gas constant

$V=\dfrac{nRT}{P}\\\\V=\dfrac{2.3\times 0.0821\times 300}{0.987 }\\\\V=57.39\ L$

So, the volume of the Ne is 57.39 L.

3 0

2 years ago