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Romashka-Z-Leto [24]
3 years ago
5

Using the average atomic masses given inside the front cover of this book, calculate the number of atoms present in each of the

following sample. a. 1.50 g of silver, Ag
b. 0.0015 mole of copper, Cu
C. 0.0015 g of copper, Cu
d. 2.00 kg of magnesium, Mg
e. 2.34 oz of calcium, Ca
f. 2.34 g of calcium, Ca
g. 2.34 moles of calcium, Ca
Chemistry
1 answer:
alexandr1967 [171]3 years ago
6 0

Answer:

c. 0.0015 g of coopet, cu

Explanation:

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4). One mole of monoclinic sulfur at 25C was placed in a constant-pressure calorimeter whose heat capacity (C) was 1620 J/K. T
andre [41]

<u>Answer:</u> The enthalpy change of the reaction is -243 J/mol

<u>Explanation:</u>

The heat released by the reaction is absorbed by the calorimeter and the solution.

The chemical equation used to calculate the heat released follows:

q=c\times \Delta T

where,

c = heat capacity of calorimeter = 1620 J/K

\Delta T = change in temperature = 0.150^oC=0.150K   (Change remains same)

Putting values in above equation, we get:

q=1620J/K\times 0.15K=243J

<u>Sign convention of heat:</u>

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

For the given chemical reaction:

S\text{ (monoclinic)}\rightarrow S\text{ (orthorhombic)}

We are given:

Moles of monoclinic sulfur = 1 mole

  • To calculate the enthalpy change of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,  

q = amount of heat released = -243 J

n = number of moles = 1 mole

\Delta H_{rxn} = enthalpy change of the reaction

Putting values in above equation, we get:

\Delta H_{rxn}=\frac{-243J}{1mol}=-243J/mol

Hence, the enthalpy change of the reaction is -243 J/mol

8 0
3 years ago
A sample contains 16 mg of polonium-218. After 12 minutes, the sample will contain 1.0 mg of polonium-218. What is the half life
mariarad [96]

Answer:

Half-life = 3 minutes

Explanation:

Using the radioactive decay equation we can solve for reaction constant, k. And by using:

K = ln2 / Half-life

We can find half-life of polonium-218

Radioactive decay:

Ln[A] = -kt + ln [A]₀

Where:

[A] could be taken as mass of polonium after t time: 1.0mg

k is Reaction constant, our incognite

t are 12 min

[A]₀ initial amount of polonium-218: 16mg

Ln[A] = -kt + ln [A]₀

Ln[1.0mg] = -k*12min + ln [16mg]

-2.7726 = - k*12min

k = 0.231min⁻¹

Half-life = ln 2 / 0.231min⁻¹

<h3>Half-life = 3 minutes</h3>

5 0
3 years ago
What mass of aluminum is needed to produce 0.500 mole of aluminum chloride?
3241004551 [841]

Answer:  " 13.5 g Al " ;

                    →  that is:  "13.5 grams of aluminum."

<u>____________________________</u>

Explanation:

<u>____________________________</u>

<u>Note</u>: What is missing from the question is the "balanced chemical equation" for the "chemical reaction" that contains:

 The reactants:  "aluminum (Al) " ;  and "chlorine (Cl) " ;  and:

 The product:    "aluminum choloride (AlCl₃) " .

____________________________

The "balanced chemical equation" is:

____________________________

        2 Al   +   3 Cl₂   →   2 AlCl₃   ;

_____________________________

<u>Note</u>: The molecular weight of "aluminum (Al)" is:   " 26.98 g /mol " .

____________________________

So:  We call solve using a technique known as:  "dimensional analysis" :

____________________________

  0.500 mol AlCl₃ * (\frac{2mol Al}{2mol AlCl_{3} }) * (\frac{26.98g Al}{1 mol Al}) = ?

____________________________

<u>Note</u>:  The units of "mol AlCl₃" cancel out to "1' ; and:

          The  units of "mol Al" cancel out to "1" ; and we are left with:

____________________________

 " \frac{(0.500 * 2 * 26.98)}{2}   g Al ["grams of aluminum"] ;

____________________________

<u>Note</u>: We can "cancel out the "2's" ; since "2/2 = 1 " ; and we have:

 →  (0.500 * 26.98) g Al ;

    = 13.49 g Al ;

         →  Round to 3 (Three) significant figures;

         →  Since:  "0.500" has 3 (Three) significant figures:

____________________________

   =  13.5 g Al ; that is:  "13.5 grams of aluminum."

____________________________

 Hope this is helpful!  

      Best wishes to you in your academic pursuits—and within the "Brainly" community!

____________________________

3 0
3 years ago
2h3po4 + 3mg(oh)2 → mg3(po4)2 + 6h2o phosphoric acid, h3po4, is neutralized by magnesium hydroxide, mg(oh)2, according to the eq
Rina8888 [55]
The balanced equation for the neutralisation reaction is as follows
2H₃PO₄ + 3Mg(OH)₂ --> Mg₃(PO₄)₂ + 6H₂O
stoichiometry of H₃PO₄ to H₂O is 2:6
number of H₃PO₄ moles reacted - 0.24 mol
if 2 mol of H₃PO₄ form 6 mol of H₂O
then 0.24 mol of H₃PO₄ forms - 6/2 x 0.24 = 0.72 mol of H₂O
therefore 0.72 mol of H₂O are formed 
4 0
3 years ago
What is something that is not moving with respect to an observer that can be used to detect motion
larisa86 [58]

Answer:

Frame of reference

Explanation:

3 0
3 years ago
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