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Jet001 [13]
3 years ago
5

In a fizzy drink container how is steel separated from aluminium

Chemistry
1 answer:
barxatty [35]3 years ago
5 0

Answer:

By filtration

Explanation:

In the container, pour water and shake the mixture very well.

Pour the mixture in a clean container thro<u>ugh</u> a <em><u>filt</u></em><em><u>er</u></em><em><u> </u></em><em><u>tunnel</u></em> and steel with be the residue and aluminium will be in water.

Add dilute Ammonia solution to the filtrate to precipitate it and burn until aluminium solid is obtained.

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Select the correct answer.
GREYUIT [131]

Answer:HNO3 + NaOH → H2O + NaNO3

Explanation:

8 0
4 years ago
I am a metal which can be drawn into thin wires
Gwar [14]

Answer:

Silver and copper

Explanation:

Siliver

Electrical conductivity=62000000(s/m)

Boiling point=2162

Melting point=961.78

Phase=solid

Copper

Electrical conductivity=59000000(s/m)

Boiling point =2927

Melting point=1084.62

Phase=soild

Metals like gold, silver and copper are drawn into thin wire and are used in an electric circuit because of their high electrical conductivity

Mostly Siliver and copper are used since using gold is much expensive.

3 0
3 years ago
Pentaborane-9, B5H9, is a colorless, highly reactive liquid that will burst into flame when exposed to oxygen. The reaction is 2
mina [271]

<u>Answer:</u> The amount of energy released per gram of B_5H_9 is -71.92 kJ

<u>Explanation:</u>

For the given chemical reaction:

2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(5\times \Delta H^o_f_{(B_2O_3(s))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(2\times \Delta H^o_f_{(B_5H_9(l))})+(12\times \Delta H^o_f_{(O_2(g))})]

Taking the standard enthalpy of formation:

\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol\\\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(5\times (1271.94))+(9\times (-285.83))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H^o_{rxn}=-9078.57kJ

We know that:

Molar mass of pentaborane -9 = 63.12 g/mol

By Stoichiometry of the reaction:

If 2 moles of B_5H_9 produces -9078.57 kJ of energy.

Or,

If (2\times 63.12)g of B_5H_9 produces -9078.57 kJ of energy

Then, 1 gram of B_5H_9 will produce = \frac{-9078.57kJ}{(2\times 63.12)}\times 1g=-71.92kJ of energy.

Hence, the amount of energy released per gram of B_5H_9 is -71.92 kJ

8 0
3 years ago
Which of the following is an oxidation reaction? A. AuCl4- → AuCl 2- B. Mn7+ → Mn2+ C. Co3+ → Co2+ D. Cl2 → ClO3-
OLga [1]
D. There must be oxygen added. 2 of them don't even have oxygen anywhere in the formula, and the 3rd loses oxygen, which is reduction.
6 0
3 years ago
Read 2 more answers
10
MAVERICK [17]

The equilibrium constant (K) : 11.85

<h3>Further explanation</h3>

Given

Reaction

N₂(g) + 3H₂(g) ⇒ 2NH₃(g)

Required

K(equilibrium constant)

Solution

The equilibrium constant (K) is the value of the concentration product in the equilibrium

The equilibrium constant based on concentration (K) in a reaction  

pA + qB -----> mC + nD  

\tt K=\dfrac{[C]^m[D]^n}{[A]^p[B]^q}

For the reaction above :

\tt K=\dfrac{[NH_3]^2}{[N_2][H_2]^3}\\\\K=\dfrac{0.1^2}{0.25\times 0.15^3}\\\\K=11.85

6 0
3 years ago
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