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Pachacha [2.7K]
4 years ago
7

The chemical formula for aluminum + oxygen

Chemistry
1 answer:
sineoko [7]4 years ago
6 0
Al2O3 is the formula 2 and 3 are subscripts

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Using the system of rules for naming binary acids and naming other binary covalent molecules​
ANEK [815]

Answer:

Atoms

Explanation:

This is also a form of molucules

3 0
3 years ago
Calculate the number of molecules in 6.7 moles of AlCl3
Greeley [361]

According to the Avogadro’s law I mole of any substance contains = 6.023 x 1023

Number of molecules in one mole= 6.023 x 1023

Therefore number of molecules in 6.7 moles of AlCl3:

6.023x1023 molecules /1.0 mole * 6.7 mole

= 40.3541 x 1023 molecules

= 4.03541 x 1024 molecules


3 0
4 years ago
Write the net ionic equ.
Artyom0805 [142]

Answer:

a) AgNO3 + KI → AgI + KNO3

b) Ba(OH)2 + 2HNO3 → Ba(NO3)2 + 2H2O

c) 2Na3PO4  + 3Ni(NO3)2  → Ni3(PO4)2 + 6NaNO3

d) 2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O

Explanation:

a) AgNO3 + KI → Ag+ + NO3- + K+ + I-

Ag+ + NO3- + K+ + I-  → AgI + KNO3

AgNO3 + KI → AgI + KNO3

b) Ba(OH)2 + 2HNO3 → Ba^2+ + 2OH- + 2H+ + 2NO3-

Ba^2+ + 2OH- + 2H+ + 2NO3- → Ba(NO3)2 + 2H2O

Ba(OH)2 + 2HNO3 → Ba(NO3)2 + 2H2O

c) 2Na3PO4  + 3Ni(NO3)2 → 6Na+ + 2PO4^3- + 3Ni^2+ + 6NO3-

6Na+ + 2PO4^3- + 3Ni^2+ + 6NO3- →  Ni3(PO4)2 + 6NaNO3

2Na3PO4  + 3Ni(NO3)2  → Ni3(PO4)2 + 6NaNO3

d) 2Al(OH)3 + 3H2SO4 → 2Al^3+ + 6OH- + 6H+ + 3SO4^2-

2Al^3+ + 3OH- + 3H+ + 3SO4^2- → Al2(SO4)3 + 6H2O

2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O

7 0
4 years ago
The density of a mixture measures 1015.3 kilograms per cubic meter. Use scientific notation to represent the density of the mixt
nata0808 [166]
1.0153 x 10^3

essentially, scientific notation requires you to take a very large or very small number and simplify it into the first few digits times 10 raised to the power of x.
Although realistically, there is no practical reason to simplify a number that is already that close to 1.

3 0
3 years ago
Read 2 more answers
Chloroform, formerly used as an anaesthetic and now believed to be a carcinogen, has a heat of vaporization ΔHvaporization = 31.
djverab [1.8K]

Answer:

Chloroform is expected to boil at 333 K (60 ^{0}\textrm{C}).

Explanation:

For liquid-vapor equilibrium at 1 atm, \Delta G^{0} = 0.

We know, \Delta G^{0}=\Delta H^{0}-T\Delta S^{0} , where T is temperature in kelvin scale.

Here both \Delta H^{0} and \Delta S^{0} are corresponding to vaporization process therefore T represents boiling point of chloroform.

So, 0=(31.4\times 10^{3}\frac{J}{mol})-[T\times (94.2\frac{J}{mol.K})]

or, T = 333 K

So, at 333 K (60 ^{0}\textrm{C}) , chloroform is expected to boil.

6 0
3 years ago
Read 2 more answers
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