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Anna [14]
2 years ago
13

Various data can be used to find the composition of an alloy (a metallic mixture). Show that calculating the mass % of Mg in a m

agnesium-aluminum alloy (d = 2.40 g/cm³) using each of the following pieces of data gives the same answer (within rounding):
(b) an identical sample reacting with excess aqueous HCl forms 1.38X10⁻² mol of H₂;
Chemistry
1 answer:
PolarNik [594]2 years ago
4 0

The mass percent of Mg in the alloy is 21.3 percent.

<h3>What purposes can magnesium aluminum serve?</h3>

For usage as an absorbent, anticaking agent, opacifying agent, viscosity-increasing agent, suspending agent, tablet and capsule disintegrant, and tablet binder in cosmetic and pharmaceutical applications, it is refined into a powder.

<h3>Al and Mg may they combine to make alloys?</h3>

Magnesium increases aluminum's strength through solid solution strengthening and enhances its capacity for strain hardening.

Because these alloys have the maximum strength among non-heat-treatable aluminum alloys, they are frequently utilized in structural applications.

<h3>How robust is magnesium aluminum alloy?</h3>

Magnesium alloy has a tensile strength of 220 Mpa, compared to 230 Mpa for aluminum alloy. Aluminum alloy is stronger than magnesium alloy in the same volume.

learn more about magnesium-aluminum alloy here

brainly.com/question/14685213

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A movable piston traps 0.205 moles of an ideal gas in a vertical cylinder. If the piston slides without friction in the cylinder
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Answer : The work done on the gas will be, 418.4 J

Explanation :

First we have to calculate the volume at 270°C.

PV_1=nRT

where,

P = pressure of gas = 1 atm

V_1 = volume of gas = ?

T = temperature of gas = 270^oC=273+270=543K

n = number of moles of gas = 0.205 mol

R = gas constant  = 0.0821 L.atm/mol.K

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

(1atm)\times V_1=0.205mol\times 0.0821L.atm/mol.K\times 543K

V_1=9.12L

Now we have to calculate the volume at 24°C.

PV_2=nRT

where,

P = pressure of gas = 1 atm

V_2 = volume of gas = ?

T = temperature of gas = 24^oC=273+24=297K

n = number of moles of gas = 0.205 mol

R = gas constant  = 0.0821 L.atm/mol.K

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

(1atm)\times V_2=0.205mol\times 0.0821L.atm/mol.K\times 297K

V_2=4.99L

Now we have to calculate the work done.

Formula used :

w=-p\Delta V\\\\w=-p(V_2-V_1)

where,

w = work done

p = pressure of the gas = 1 atm

V_1 = initial volume = 9.12 L

V_2 = final volume = 4.99 L

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

w=-p(V_2-V_1)

w=-(1atm)\times (4.99-9.12)L

w=4.13L.artm=4.13\times 101.3J=418.4J

conversion used : (1 L.atm = 101.3 J)

Therefore, the work done on the gas will be, 418.4 J

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What is the mole ratio of D to A in the generic chemical reaction? 4A + B --&gt; C + D​
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<h3>Answer:</h3>

Mole ratio of D to A is 1 : 4

<h3>Explanation:</h3>

We are given the generic chemical equation;

4A + B → C + D​

We are supposed to determine the mole ratio of D to A

What is mole ratio?

  • Mole ratio is the ratio of the number of moles of reactants or products in a chemical reaction.
  • We determine the mole ratio using the coefficients of reactants or products in question.

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  • In the equation, 4A + B → C + D​, the coefficient of A is 4 while the coefficient of D is 1.
  • This means, 4 moles of A reacts with 1 mole of b to produce 1 mole of C and 1 mole of D
  • Thus, mole ratio of D to A  is 1 : 4
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