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PtichkaEL [24]
3 years ago
8

A student has a piece of aluminum metal. Which is the most reasonable assumption the student could make about the metal?

Chemistry
2 answers:
ollegr [7]3 years ago
6 0

Answer:

E. It could be stretched into a thin wire.

Explanation:

Metals are malleable. This means that they can be hammered or pressed into different shapes without braking. They are also ductile, which means that they can be drawn out into thin wires without braking.

igor_vitrenko [27]3 years ago
3 0

Aluminium belongs to 13th group of periodic table. It undergoes oxidation to given Al^+3 .

It is observed that when aluminium is added to a solution of copper sulphate the colour of the solution changes from blue to grey. It is due to formation of grey coloured solution of aluminium sulphate as

2Al^+3  + 3SO4^-2  ---> Al2(SO4)3


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Titanium is a transition metal used in many alloys because it is extremely strong and lightweight. Titanium tetrachloride (TiCl4
vova2212 [387]

Answer:

a) 226.6 grams of Cl₂

b) 19.2 grams of C

c) 303.2 grams of TiCl₄ and 70.4 grams of CO₂

Explanation:

The balanced chemical reaction is the following:

TiO₂(s) + C(s) + 2 Cl₂(g) → TiCl₄(s) + CO₂(g)

(a) What mass of Cl₂ gas is needed to react with 1.60 mol TiO₂?

From the chemical equation, 1 mol of TiO₂ reacts with 2 moles of Cl₂. So, the stoichiometric ratio is 2 mol Cl₂/1 mol TiO₂. We multiply this ratio by the moles of TiO₂ we have to calculate the moles of Cl₂ we need:

1.60 mol TiO₂ x 2 mol Cl₂/1 mol TiO₂ = 3.2 mol Cl₂

Now, we convert from moles to mass by using the molecular weight (MW) of Cl₂:

MW(Cl₂) = 35.4 g/mol x 2 = 70.8 g/mol

mass of Cl₂= 3.2 mol x 70.8 g/mol = 226.6 g

<em>Therefore, 226.6 grams of Cl₂ are needed to react with 1.6 mol of TiO₂. </em>

(b) What mass of C is needed to react with 1.60 mol of TiO₂?

From the chemical equation, 1 mol of TiO₂ reacts with 1 moles of C(s). So, the stoichiometric ratio is 1 mol C/1 mol TiO₂. We multiply this ratio by the moles of TiO₂ we have to calculate the moles of C(s) we need:

1.60 mol TiO₂ x 1 mol C(s)/1 mol TiO₂ = 1.60 mol C(s)

So, we convert the moles of C(s) to grams as follows:

MW(C) = 12 g/mol

1.60 mol x 12 g/mol = 19.2 g C(s)

<em>Therefore, a mass of 19.2 grams of C is needed to react with 1.60 mol of TiO₂. </em>

(c) What is the mass of all the products formed by reaction with 1.60 mol of TiO₂?

From the chemical equation, we can notice that 1 mol of TiO₂ produces 1 mol of TiCl₄ and 1 mol of CO₂. So, from 1.60 moles of TiO₂, 1 mol of each product will be produced:

1 mol TiO₂/1 mol TiCl₄ ⇒ 1.60 mol TiO₂/1.60 mol TiCl₄

1 mol TiO₂/1 mol CO₂ ⇒ 1.60 mol TiO₂/1.60 mol CO₂

Finally, we convert the moles to grams by using the molecular weight of each compound:

MW(TiCl₄) = 47.9 g/mol Ti + (35.4 g/mol x 4 Cl) = 189.5 g/mol

1.60 mol x 189.5 g/mol = 303.2 g

MW(CO₂) = 12 g/mol C + (16 g/mol x 2 O) = 44 g/mol

1.60 mol x 44 g/mol = 70.4 g

<em>Therefore, from the reaction of 1.60 mol of TiO₂ are formed 303.2 grams of TiCl₄ and 70.4 grams of CO₂.</em>

3 0
3 years ago
What is meaning of trash​
salantis [7]

Answer:

waste or discarded material

7 0
3 years ago
Read 2 more answers
For the following reaction, find the value of Q and predict the direction of change, given that a 1L flask initially contains 2
Tresset [83]

Answer:

C) Q < K, reaction will make more products

Explanation:

  • 1/8 S8(s)  + 3 F2(g)  ↔  SF6(g)

∴ Kc = 0.425 = [ SF6 ] / [ F2 ]³

∴ Q = [ SF6 ] / [ F2 ]³

∴ [ SF6 ] = 2 mol/L

∴ [ F2 ] = 2 mol/L

⇒ Q = ( 2 ) / ( 2³)

⇒ Q = 0.25

⇒ Q < K, reaction will make more products

 

5 0
3 years ago
The conversation of cyclopropane to propene in the gas phase is a first order reaction with a rate constant of 6.7x10-⁴s-¹. a) i
Rus_ich [418]

Answer: a)  The concentration after 8.8min is 0.17 M

b) Time taken for the concentration of cyclopropane to decrease from 0.25M to 0.15M is 687 seconds.

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process  

a) concentration after 8.8 min:

8.8\times 60s=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{a-x}

\log\frac{0.25}{a-x}=0.15

\frac{0.25}{a-x}=1.41

(a-x)=0.17M

b) for concentration to decrease from 0.25M to 0.15M

t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{0.15}\\\\t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\times 0.20

t=687s

7 0
3 years ago
How many valance electrons are there in group 10?
poizon [28]

there are valence electrons 3 in group 10.

5 0
3 years ago
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