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LiRa [457]
3 years ago
5

Solid aluminum metal and diatomic bromine liquid react spontaneously to form a solid product. give the balanced chemical equatio

n (including phases) that describes this reaction. indicate the phases using abbreviation (s), (l), or (g) for solid, liquid, or gas, respectively.
Chemistry
2 answers:
Digiron [165]3 years ago
4 0

Explanation:

Aluminium being metal reacts violently with liquid bromine. It is a highly exothermic reaction during which large amount of heat is released.

Aluminum when reacts with liquid bromine it gives aluminium bromide as a product.

The balanced chemical equation is given as:

2Al(s) + 3Br_2(l)\rightarrow 2AlBr_3(s)

According stoichiometry, 2 moles of aluminum metal react with 3 mole of liquid bromine to give 2 moles of aluminum bromide.

kaheart [24]3 years ago
3 0
Aluminum has a chemical formula of Al, while diatomic bromine has a chemical formula of Br₂. The balanced chemical reaction is shown below:

<em>2 Al (s) + 3 Br₂ (l) → 2 AlBr₃ (s)</em>

The solid product is called Dibromoaluminum. The stoichiometric coefficients are used to balance the reaction to obey the Law of Conservation of Mass.
You might be interested in
when 45 grams of copper (ii) carbonate are decomposed with heat how many grams of carbon dioxide will be produced
Maksim231197 [3]

Answer:

16.02 g

Explanation:

the balanced equation for the decomposition of CuCO₃ is as follows

CuCO₃ --> CuO + CO₂

molar ratio of CuCO₃ to CO₂ is 1:1

number of CuCO₃ moles decomposed - 45 g / 123.5 g/mol = 0.364 mol

according to the molar ratio

1 mol of CuCO₃ decomposes to form 1 mol of CO₂

therefore 0.364 mol of  CuCO₃ decomposes to form 0.364 mol of CO₂

number of CO₂ moles produced - 0.364 mol

therefore mass of CO₂ produced - 0.364 mol x 44 g/mol = 16.02 g

16.02 g of CO₂ produced

8 0
3 years ago
If element x forms the oxides xo and x2o3 the oxidation numbers of element x are
Deffense [45]

Answer:

b) +2 and +3.

Explanation:

Hello,

In this case, given the molecular formulas:

XO

And:

X_2O_3

We can relate the subscripts with the oxidation states by knowing that they are crossed when the compound is formed, for that reason, we notice that oxygen oxidation state should be -2 for both cases and the oxidation state of X in the first formula must be +2 since both X and O has one as their subscript as they were simplified:

X^{+2}O^{-2}

Moreover, for the second case the oxidation state of X should be +3 in order to obtain 3 as the subscript of oxygen:

X_2^{+3}O_3^{-2}

Thus, answer is b)+2 and +3

Best regards.

3 0
2 years ago
HELPPPPPPPPP ASAPPPP!!!!!
andreev551 [17]
Answer :
B

!!!!!!!!!!!!!!!
8 0
3 years ago
1.25 cm is the same distance as ?<br> 12.5 Km<br> 12.5 mm<br> 12.5 dm<br> 12.5 m
pogonyaev

Answer:

12.5mm

Explanation:

1cm = 10mm

so you need to multiply the 1.25 by 10 to get it in mm.

6 0
3 years ago
Calculate the molality of isoborneol in the product if, a) the melting point of pure camphor is 179°C and the melting point take
tresset_1 [31]

Answer:

The molality of isoborneol in camphor is 0.53 mol/kg.

Explanation:

Melting point of pure camphor= T =179°C

Melting point of sample = T_f = 165°C

Depression in freezing point = \Delta T_f=?

\Delta T_f=T- T_f=179^oC-165^oC=14^oC

Depression in freezing point  is also given by formula:

\Delta T_f=i\times K_f\times m

K_f = The freezing point depression constant

m = molality of the sample

i = van't Hoff factor

We have: K_f = 40°C kg/mol

i = 1 (  organic compounds)

\Delta T_f=14^oC

14^oC=1\times 40^oC kg/mol\times m

m=\frac{14^oC}{1\times 40^oC kg/mol}=0.35 mol/kg

The molality of isoborneol in camphor is 0.53 mol/kg.

8 0
2 years ago
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