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gayaneshka [121]

2 years ago

11

when magnesium is burned in air the mass of the product is greater than the mass of the solid magnesium burned. A student claims

that this is a violation of the Low of conservation of mass do you agree or disagree with the claim. justify your answer

2 answers:

Nikitich [7]2 years ago

5 0

Veg egverbygbe rbbnfghjufddfg

il63 [147K]2 years ago

4 0

No, this is not a violation of the Law of conservation because when Magnesium is burnt it reacts with oxygen to form magnesium oxide. This end product is going to weigh more than the solid magnesium due to the addition of oxygen.

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How many liters of oxygen are required to react completely with 14.8 mol of Al?

Bond [772]

Answer:

296 L

Explanation:

We will need a balanced equation with moles, so let's gather all the information in one place.

4Al + 3O₂ ⟶ 2Al₂O₃

n/mol: 17.4

1. Moles of O₂

$n = \text{17.4 mol Al}\times \dfrac{\text{3 mol O}_{2}}{\text{4 mol Al}}= \text{13.05 mol O}_{2}$

2. Volume of O₂

You haven't given the conditions at which the volume is measured, so I assume it is at STP (0 °C and 1 bar).

At STP the molar volume of a gas is 22.71 L.

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3 years ago

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Anna11 [10]

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3 years ago

A 50.0 mL sample of a 1.00 M solution of CuSO4 is mixed with 50.0 mL of 2.00 M KOH in a calorimeter. The temperature of both sol

Reika [66]

Answer : The enthalpy change for the process is 52.5 kJ/mole.

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the solution

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the solution

$c_1$ = specific heat of calorimeter = $12.1J/^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_2$ = mass of water or solution = $Density\times Volume=1/mL\times 100.0mL=100.0g$

$\Delta T$ = change in temperature = $T_2-T_1=(26.3-20.2)^oC=6.1^oC$

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

$q=[(12.1J/^oC\times 6.1^oC)+(100.0g\times 4.18J/g^oC\times 6.1^oC)]$

$q=2623.61J$

Now we have to calculate the enthalpy change for the process.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat released = 2626.61 J

n = number of moles of copper sulfate used = $Concentration\times Volume=1M\times 0.050L=0.050mole$

$\Delta H=\frac{2623.61J}{0.050mole}=52472.2J/mole=52.5kJ/mole$

Therefore, the enthalpy change for the process is 52.5 kJ/mole.

8 0

3 years ago