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4 years ago

In the following chemical reaction, what product is represented by X?

Chemistry

1 answer:

olasank [31]4 years ago

6 0

Answer:

Al (Aluminum)

Al (OH)₃

Explanation:

Reaction Given:

AlCl₃ + NaOH -------> X (OH)₃ + NaCl

As we see at the reactants there are to compounds combining together

that is Aluminum Chloride and Sodium hydroxide.

Reaction Reactant:

AlCl₃

NaOH

This reaction is type of double displacement.

Now

if we look at the product side we again have 2 types of products

Reaction Products

1. X(OH)₃

2. NaCl

So,

Also according to law of definite composition a chemical compound have fixed ratio of component in product.

So, If we look at both the reactant and product atoms in compounds only Aluminum (Al) is missing on the product side.

Also if we look at the product i.e. X(OH)₃ means that 3 OH groups are combine with an atom that have the tendency of combining with 3 OH group and have valency of 3.

So only aluminum have the 3+ valency and can form a compound with OH have 3 at its base in unit formula.

So,

The product will be

Al (OH)₃

And the balanced chemical reaction will be

AlCl₃ + 3NaOH -------> Al(OH)₃ + 3NaCl

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A 1.93-mol sample of xenon gas is maintained in a 0.805-L container at 306 K. Calculate the pressure of the gas using both the i

Alex

Answer : The pressure of the gas using both the ideal gas law and the van der Waals equation is, 60.2 atm and 44.6 atm respectively.

Explanation :

First we have to calculate the pressure of gas by using ideal gas equation.

$PV=nRT$

where,

P = Pressure of $Xe$ gas = ?

V = Volume of $Xe$ gas = 0.805 L

n = number of moles $Xe$ = 1.93 mole

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $Xe$ gas = 306 K

Now put all the given values in above equation, we get:

$P\times 0.805L=1.93mole\times (0.0821L.atm/mol.K)\times 306K$

$P=60.2atm$

Now we have to calculate the pressure of gas by using van der Waals equation.

$(P+\frac{an^2}{V^2})(V-nb)=nRT$

P = Pressure of $Xe$ gas = ?

V = Volume of $Xe$ gas = 0.805 L

n = number of moles $Xe$ = 1.93 mole

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $Xe$ gas = 306 K

a = pressure constant = $4.19L^2atm/mol^2$

b = volume constant = $5.11\times 10^{-2}L/mol$

Now put all the given values in above equation, we get:

$(P+\frac{(4.19L^2atm/mol^2)\times (1.93mole)^2}{(0.805L)^2})[0.805L-(1.93mole)\times (5.11\times 10^{-2}L/mol)]=1.93mole\times (0.0821L.atm/mol.K)\times 306K$

$P=44.6atm$

Therefore, the pressure of the gas using both the ideal gas law and the van der Waals equation is, 60.2 atm and 44.6 atm respectively.

5 0

3 years ago

A sample of seawater contains 78.0 grams of NaCl in 2,025 mL of solution. If the molar mass of NaCl is 58.443 g/mol, what is the

DENIUS [597]

Answer:

that would be 0.659 M NaCl

4 0

3 years ago

Read 2 more answers

A sample of helium gas has a pressure of 1.20 atm at 22.0 C. At what Celsius temperature will the helium reach a pressure of 2.0

Anarel [89]

Answer: 36.6°C

Explanation:

Given that,

initial pressure of helium (P1) = 1.20 atm

Initial temperature (T1) = 22.0°C

Final temperature (T2) = ?

Final pressure of helium (P2) = 2.00 atm

Since pressure and temperature are given while volume is constant, apply the formula for pressure's law

P1/T1= P2/T2

1.20 atm / 22.0°C = 2.00 atm / T2

Cross multiply

1.20 atm•T2= 2.00 atm•22°C

1.20 atm•T2= 44 atm•°C

Divide both sides by 1.20 atm

1.20 atm•T2/1.20 atm = 44 atm•°C/1.20 atm

T2 = 36.6°C

8 0

4 years ago

which activity most contributed to economic interdependence among West African peoples? farming, hurding, textiles, trade.

olga_2 [115]

Tradeing
Explanation:

7 0

3 years ago

How many grams of chromium are required to react with 125ml of 0.75m cuso4?

evablogger [386]

Chromium and copper (ii) sulfate react according the reaction below;
2Cr + 3CuSO4 = 3Cu +Cr2(SO4)3
We can calculate the number of moles of CuSO4 present;
no of moles of CuSO4 = 0.75 × 0.125 = 0.09735 moles
According to the reaction the mole ratio is 2 : 3
Moles of Chromium = (0.09735 × 2)/3
= 0.0625 Moles
Mass of chromium = 0.0625 moles × 52 g/mole
= 3.25 g of chromium

8 0

3 years ago