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ollegr [7]
3 years ago
10

Argue whether this chemical reaction supports or does not support the Law of Conservation of Matter.

Chemistry
1 answer:
Anit [1.1K]3 years ago
4 0

Explanation:

The reaction is as follows:

2Mg(s) + O2(g) ---> 2MgO(s)

and the researcher said that 32 g of MgO was produced.

Stoichiometry:

28 g Mg × (1 mol Mg/24.305 g Mg) = 1.15 mol Mg

15 g O2 × (1 mol O2/15.999 g O2) = 0.938 mol O2

1.15 mol Mg × (2 mol MgO/2 mol MgO) = 1.15 mol MgO

1.15 mol MgO × (40.3044 g MgO/1 mol MgO) = 46.6 g MgO

0.938 mol O2 × (2 mol MgO/1 mol O2) = 1.88 mol MgO

1.88 mol MgO × (40.3044 g MgO/1 mol MgO = 75.6 g MgO

Based on these numbers, the amount of product after the reaction is much less than expected so these results don't seem to support the law of conservation of matter.

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A Bronsted-Lowry<br> _______is a molecule or ion that donates<br> a hydrogen ion in a reaction,
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3 years ago
For the reaction, a → b, the rate constant is 0.0208 m-1 sec-1. How long would it take for [a] to decrease from 0.100 to 0.0450
goldfiish [28.3K]

First, assume the order of the given reaction is n, then the rate of reaction i.e. \frac{dx}{dt}=k\times[A]^{n}

where, dx is change in concentration of A in small time interval dt and k is rate constant.

According to units of rate constant, the reaction is of second order.

\frac{1}{[A]_{t}}-\frac{1}{[A]_{o}} = kt   (second order formula)

Put the values,

\frac{1}{0.04590 m}-\frac{1}{0.100 m} =0.0208 m^{-1}s^{-1} \times t  

 22.23 m -10 m =0.0208 m^{-1}s^{-1} \times t

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3 years ago
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When a metal was exposed to light at a frequency of 4.07× 1015 s–1, electrons were emitted with a kinetic energy of 3.30× 10–19
Licemer1 [7]

Answer :  The maximum number of electrons released = 1.432\times 10^{12}electrons

Explanation : Given,

Frequency = 4.07\times 10^{15}s^{-1}

Kinetic energy = 3.30\times 10^{-19}J

Total energy = 3.39\times 10^{-7}J

First we have to calculate the work function of the metal.

Formula used :

K.E=h\nu -w

where,

K.E = kinetic energy

h = Planck's constant = 6.626\times 10^{-34}J/s

\nu = frequency

w = work function

Now put all the given values in this formula, we get the work function of the metal.

3.30\times 10^{-19}J=(6.626\times 10^{-34}J/s\times 4.07\times 10^{15}s^{-1})-w

By rearranging the terms, we get

w=2.367\times 10^{-18}J

Therefore, the works function of the metal is, 2.367\times 10^{-18}J

Now we have to calculate the maximum number of electrons released.

The maximum number of electrons released = \frac{\text{ Total energy}}{\text{ work function}}

The maximum number of electrons released = \frac{3.39\times 10^{-7}J}{2.367\times 10^{-19}J}=1.432\times 10^{12}electrons

Therefore, the maximum number of electrons released is 1.432\times 10^{12}electrons

8 0
3 years ago
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