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garri49 [273]
4 years ago
15

8. A 10.0-g sample of magnesium reacts with oxygen to form 16.6 g of magnesium oxide. How many grams of oxygen reacted? 9. CHALL

ENGE 106.5 g of HCl(g) react with an unknown amount of NH (9) to produce 156.3 g of NH,Cl(s). How many grams of NH (9) reacted? Is the law of conservation of mass observed in the reaction? Justify your answer.
Chemistry
1 answer:
Veronika [31]4 years ago
3 0

Answer:

8. 13.3g of Oxygen is reacted with 10.0g of Magnesium

9. 49.6g of Ammonia (NH₃) reacts with 106.5g of HCl

   The law of conservation of mass is observed in the reaction

Explanation:

8. The Relative Atomic Masses of the elements involved are:  

   Mg=24, O=16

   The equation for the reaction is given as  

   2Mg(s) + O₂(g) → 2MgO(s)  

   From the equation above

   Let's assume that the amount in gram of Oxygen required to react with 10.0g of Mg is <em>x</em>

   2(24)g of Magnesium is required to react with (16 x 2)g of Oxygen

   10.0g of Magnesium should react with <em>x</em> g of Oxygen  

   ∴ <em>x</em> = (10.0 x (16 x 2))/2(24) = (10.0 x 32)/48 = 13.3g of Oxygen

9.  The Relative Atomic Masses of the elements involved in the reaction are  

    H=1, Cl=35.5, N=14  

    The equation for the reaction is given as  

    HCl(g) + NH₃(g) → NH₄Cl(s)  

    From the equation above

    Let's say <em>y</em> equals the amount in grams of Ammonia involved in the reaction  

    (1 + 35.5)g of HCl reacts with (14+(1 x 3))g of Ammonia  

    106g of HCl should react with <em>y</em> g of Ammonia

∴ <em>y</em> = (106 x (14 + (1 x 3)))/(1+35.5) = (106 x 17)/36.5 = 49.6g of Ammonia

The  law of conservation of mass states that the total mass in a closed system (reaction) remains constant over time (i.e mass is neither gained or lost). In other words, the total mass of the reactant equals the total mass of the product.

Based on the reaction above, the total mass of the reactant (i.e HCl and NH₃) is 53.5g which is the same as the product (i.e NH₄Cl) which is also 53.5g.

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PLZ HELP ...The half-life of polonium-218 is 3.0 minutes. If you start with 40.0 g, how long will it be before only 4.5 g remain
Elanso [62]

Answer:-  9.4 minutes.

Solution:- Radioactive decay obeys first order reaction kinetics and the equation used to solve this type of problems is:

lnN=-kt+lnN_0

where, k is decay constant and t is the time. N_0 is the initial amount of the radioactive substance and N is the remaining amount.

Since the value of decay constant is not given, so we need to calculate it first from given half life by using the formula:

k=\frac{0.693}{t_1_/_2}

where t_1_/_2 stands for half life.

Given half life is 3.0 minutes.

So, k=\frac{0.693}{3.0min}

k=0.231min^-^1

Let's plug in the values in the first order reaction equation and solve it for t.

ln4.5g=-0.231min^-^1(t)+ln40.0g

It could also be written as:

ln(\frac{4.5g}{40.0g})=-0.231min^-^1}

-2.18=-0.231min^-^1}

t=\frac{-2.18}{-0.231min^-^1}

k = 9.4 min

So, the radioactive substance would take 9.4 minutes to decay from 40.0 grams to 4.5 grams.

4 0
3 years ago
In class we derived the Gibbs energy of mixing for a binary mixture of perfect gases. We also discussed that the same result is
GaryK [48]

Answer:

Attached below

Explanation:

Free energy of mixing = ΔGmix = Gf - Gi

attached below is the required derivation of the

<u>a) Molar Gibbs energy of mixing</u>

ΔGmix = Gf - Gi

hence : ΔGmix = ∩RT ( X1 In X1 + X2 In X2 + X3 In X3 + ------- )

<u>b) molar excess Gibbs energy of mixing</u>

Ni = chemical potential of gas

fi = Fugacity

N°i = Chemical potential of gas when Fugacity = 1

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3 years ago
How many oxygen atoms are present in 23.6 grams of aspartame?
olga55 [171]
The formula of aspartame is <span>C14H18N2O5.
From the periodic table:
molecular mass of hydrogen = 1 grams
molecular mass of carbon = 12 grams
molecular mass of nitrogen = 14 grams
molecular mass of oxygen = 16 grams

This means that:
molar mass of aspartame = 14(12)+18(1)+2(14)+5(16) = 294 grams
Therefore, each 294 grams of aspartame contains 5(16) = 80 grams of oxygen. To know number of grams of oxygen in 23.6 grams of aspartame, simply use cross multiplication as follows:
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7 0
3 years ago
Shane performed the following trials in an experiment. Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0
Strike441 [17]

Answer:

A. The same amount of heat is absorbed in both the experiments because the product of mass, specific heat capacity, and change in temperature are equal for both.

Explanation:

  • The amount of heat absorbed by water (Q) can be calculated from the relation:

<em>Q = m.c.ΔT.</em>

where, Q is the amount of heat absorbed by water,

m is the mass of water,

c is the specific heat capacity of water (c = 4.186 J/g °C),

ΔT is the temperature difference (final T - initial T).

  • <u><em>For trial 1:</em></u>

m = 30.0 g, c = 4.18 J/g °C, ΔT = 40.0 °C – 0.0 °C = 40.0 °C

<em>∴ Q = m.c.ΔT</em> = (30.0 g)(4.18 J/g °C)(40.0 °C) = <em>5023 J.</em>

  • <u><em>For trial 2:</em></u>

m = 40.0 g, c = 4.18 J/g °C, ΔT = 40.0 °C – 10.0 °C = 30.0 °C

<em>∴ Q = m.c.ΔT</em> = (40.0 g)(4.18 J/g °C)(30.0 °C) = <em>5023 J.</em>

  • So, the right choice is:

<em>A. The same amount of heat is absorbed in both the experiments because the product of mass, specific heat capacity, and change in temperature are equal for both. </em>

4 0
3 years ago
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