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olya-2409 [2.1K]
3 years ago
10

A sample of an alloy of aluminum contains 0.0898 mol Al and 0.0381 mol Mg. What are the mass percentages of Al and Mg in the all

oy?
Chemistry
2 answers:
blondinia [14]3 years ago
7 0

Answer:

Al 72.61%

Mg 27.39%

Explanation:

To obtain the mass percentages, we need to place the individual masses over the total mass and multiply by 100%.

If we observe clearly, we can see that the parameters given are the moles. We need to convert the moles to mass.

To do this ,we need to multiply the moles by the atomic masses. The atomic mass of aluminum is 27 while that of magnesium is 24.

Now, the mass of aluminum is thus = 27 * 0.0898 = 2.4246g

The mass of magnesium is 0.0381 * 24 = 0.9144g

We can now calculate the mass percentage.

The total mass is 0.9144 + 2.4246 = 3.339g

% mass of Al = 2.4246/3.339 * 100 = 72.61%

% mass of Mg = 0.9144/3.39 * 100 = 27.39%

marshall27 [118]3 years ago
4 0

Answer:

Mass % Al = 72.3 %

Mass % Mg = 27.7 %

Explanation:

Step 1: Data given

Number of moles Al = 0.0898 moles

Number of moles Mg = 0.0381 moles

Molar mass Al = 26.98 g/mol

Molar mass Mg = 24.3 g/mol

Step 2: Calculate mass Al

Mass Al = moles Al * molar mass Al

Mass Al = 0.0898 moles * 26.98 g/mol

Mass Al = 2.42 grams

Step 3: Calculate mass Mg

Mass Mg = 0.0381 moles * 24.3 g/mol

Mass Mg = 0.926 grams

Step 4: Calculate total mass

Total mass = mass Al + mass Mg

Total mass = 2.42 grams + 0.926 grams

Total mass = 3.346 grams

Step 5: Calculate mass %

Mass % Al = (mass Al/ total mass) * 100%

Mass % Al = (2.42 grams / 3.346 grams ) *100%

Mass % Al = 72.3 %

Mass % Mg = (mass Mg/ total mass)*100%

Mass % Mg = (0.926 / 3.346) *100 %

Mass % Mg = 27.7 %

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Answer:

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Explanation:

When water reacts with formic acid, The following equilibrium is set up;

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This is because, the water abstracts a proton from formic acid to form its conjugate base, formate ion.

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Joan’s initial nickel (II) chloride sample was green and weighed 4.3872 g. After the dehydration reaction and removal of excess
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Answer:

a) yes, it was an hydrate

b) the number of waters of hydration, x = 6

Explanation:

a) yes it was an hydrate because the mass decreased after the process of dehydration which means removal of water thus some water molecules were present in the sample.

b) NiCl2. xH2O

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mass of water in the hydrated sample = mass of hydrated - mass of dehydrated = 4.3872 - 2.3921 = 1.9951 g which represent the mass of water that was present in the hydrated sample.

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Divide both by the smallest number of mole (which is for NiCl2) to find the coefficient of each

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3 years ago
A 45-g aluminium spoon(specific heat 0.80 / J/gdegree Celsius) at 24 degree celsius is placed in 180 ml(180 grams) of coffee at
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Explanation:

a) The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, heat_{absorbed}=heat_{released}

To calculate the amount of heat released or absorbed, we use the equation:  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

Also,

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]    ..........(1)

where,

q = heat absorbed or released

m_1 = mass of aluminium = 45 g

m_2 = mass of coffee = 180 g

T_{final} = final temperature = ?

T_1 = temperature of aluminium = 24^oC

T_2 = temperature of coffee = 85^oC

c_1 = specific heat of aluminium = 0.80J/g^oC

c_2 = specific heat of coffee= 4.186 J/g^oC

Putting all the values in equation 1, we get:

45 g\times 0.80J/g^oC\times (T_{final}-24^oC)=-[180 g\times 4.186J/g^oC\times (T_{final}-83^oC)]

T_{final}=80.30^oC

80.30 °C is the final temperature.

b) Energy flows from higher temperature to lower temperature.Whenever two bodies with different energies and temperature come in contact. And the resulting temperature of both bodies will less then the body with high temperature and will be more then the body with lower temperature.

So, is our final temperature of both aluminium and coffee that is 80°C less than initial temperature of coffee and more than the initial temperature of the aluminum.

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