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3 years ago

What mass of HCL, in grams, is required to react with 0.610 g of al(oh)3 ?

Chemistry

1 answer:

kompoz [17]3 years ago

3 0

Answer: 0.8541 grams of HCl will be required.

Explanation: Moles can be calculated by using the formula:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of $Al(OH)_3$ = 0.610 g

Molar mass of $Al(OH)_3$ = 78 g/mol

$\text{Number of moles}=\frac{0.610g}{78g/mol}$

Number of moles of $Al(OH)_3$ = 0.0078 moles

The reaction between $Al(OH)_3$ and HCl is a type of neutralization reaction because here acid and base are reacting to form an salt and also releases water.

Chemical equation for the above reaction follows:

$Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O$

By Stoichiometry,

1 mole of $Al(OH)_3$ reacts with 3 moles of HCl

So, 0.0078 moles of $Al(OH)_3$ will react with $\frac{3}{1}\times 0.0078$ = 0.0234 moles

Mass of HCl is calculated by using the mole formula, we get

Molar mass of HCl = 36.5 g/mol

Putting values in the equation, we get

$0.0234moles=\frac{\text{Given mass}}{36.5g/mol}$

Mass of HCl required will be = 0.8541 grams

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Answer:

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Explanation:

Given:

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Using ideal gas equation,

P × V = n × R × T

But n = mass/molar mass

V = (R × T)/P

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g All of the molecules below have polar bonds but only one of the molecules is a polar molecule. Which one is a polar molecule?

lbvjy [14]

Answer: $NF_3$

Explanation:

Geometrical symmetry of the molecule and the polarity of the bonds determine the polarity of the molecule.

The molecule that has zero dipole moment that means it is a geometrically symmetric molecule and the molecule which has some net dipole moment means it is a geometrically asymmetric molecule.

As the molecule is symmetric, the dipole moment will be zero as dipole moments cancel each other and the molecule will be non-polar.

As the molecule is asymmetric, the dipole moment will not be zero and the molecule will be polar.

Example: $NF_3$

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3 years ago

Three 6−l flasks, fixed with pressure gauges and small valves, each contain 6 g of gas at 276 k. flask a contains ch4, flask b c

Margaret [11]

First, please check the missing part in your question in the attachment.

a) So first, the Rank of pressure:

according to this formula PV = nRT and when n = m/Mw

PV = m/Mw * R*T

when we have the same mass m and the same V volume so P will proportional with the mole weight M as when the M is smaller the pressure will be greater

when Mw of H2(A) = 2 g / Mw of He (B) = 4 g and Mw of CH4(C) = 16 g

∴ Pressure :

(A) > (B) > (c)

B) The rank of average molecular kinetic energy:

when K = 3/2 KB T

when K is the average kinetic energy per molecule of gas

and KB is Boltzmann's constant

and T is the temperature (K)

So from this equation, we can know that K only depends on T value, and when we have the T constant here for A, B, and C So the rank of K will be like the following:

∴ A = B = C

C) the rank of diffusion rate after the valve is opened:

according to this formula:

R2/R1 = √M1/M2

from this equation, we can see that diffusion is proportional to the reciprocal of the molecular mass M so,

when Mw H2 (A) = 2 g & Mw He(B) = 4 g & CH4 (C) = 16 g

∴ the rank of diffusion:

A > B > C

D) The rank of the Total kinetic energy of the molecules:

when we have the Mw different so it will make the no.of molecules differs as when the Mw is low the no.of molecules will be hight, and when the average molecular kinetic energy equals. so the total kinetic energy will depend on no. of molecules

∵ Mw A < Mw B < Mw C

∴no .of molecules of A > B >C

∴ the rank of total kinetic energy is:

A > B > C

e) the rank of density:

when ρ = m/ v

and m is the mass & v is the volume and we have both is the same for A, B, and C

so the density also will be the same, ∴ the rank of the density is:

A = B = C

F) the rank of the collision frequency:

as the no.of molecules increase the collision frequency increase and depend also on the velocity and it's here the same.

∴ Collision frequency will only depend on the no.of molecules

we have no.of molecules of A > B > C as Mw A < B < C

∴the rank of the collision frequency is:

A > B > C

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