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Elan Coil [88]
3 years ago
7

Copper has two naturally occurring isotopes. Cu−63 has a mass of 62.939 amu and relative abundance of 69.17%.

Chemistry
1 answer:
fiasKO [112]3 years ago
7 0
The answer is 64.907 amu.

The atomic mass of an element is the average of the atomic masses of its isotopes. The relative abundance of isotopes must be taken into consideration, therefore:
atomic mass of copper = atomic mass of isotope 1 * abundance 1 + atomic mass of isotope 2 * abundance 2

We know:
atomic mass of copper = 63.546 amu
The atomic mass of isotope 1 is: 62.939 amu
The abundance of isotope 1 is: 69.17% = 0.6917
The atomic mass of isotope 1 is: x
The abundance of isotope 2: 100% - 69.17% = 30.83% = 0.3083

Thus:
63.546 amu = 62.939 amu * 0.6917 + x * 0.3083
63.546 <span>amu = 43.535 amu + 0.3083x
</span>⇒ 63.546 amu - 43.535 amu = 0.3083x
⇒ 20.011 amu = 0.3083x
   ⇒ x = 20.011 amu ÷ 0.3083 = 64.907 amu
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Mekhanik [1.2K]

Answer: 1.80g

Explanation:

Molar Mass of AlCl3 = 27 + (3x35.5)

= 27 + 106.5 = 133.5g/mol

Number of mole of AlCl3 = 0.0135mol

Mass = 0.0135 x 133.5= 1.80g

6 0
3 years ago
2) A professor wanted to set up a similar experiment as the one you performed in lab. He wanted to use Al(OH)3 in place of Ca(OH
expeople1 [14]

Explanation:

The only flaw I can find is you squared 3 instead of cubing it and it will be 27X^4 instead of 9x^4.

This reduces the amount slightly, but the number is still incredibly high (about 10 ^ 5 L is what I've calculated). Your professor might want to point out that this will not be a effective experiment due to the large volume of saturated

The Ksp value of Ca(OH)2 on the site (I used 5.5E-6 [a far more soluble compound than Al(OH)3]) and estimated how much of it will be needed. My calculation was approximately 30 ml. If you were using that much in the experiment, it implies so our estimates for Al(OH)3 are right, that the high amount is unreasonably big and that Al(OH)3 will not be a suitable replacement unless the procedure was modified slightly.

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3 years ago
C2h6o how many moles of ethanol are present in a 10.0 g sample of ethanol
Dmitriy789 [7]
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5 0
3 years ago
Read 2 more answers
When 100 mL of 0.200 M NaCl(aq) and 100 mL of 0.200 M AgNO3(aq), both at 21.9 °C, are mixed in a coffee cup calorimeter, the tem
masya89 [10]

Answer:

There is 1.3 kJ heat produced(released)

Explanation:

<u>Step 1:</u> Data given

Volume of a 0.200 M Nacl solution = 100 mL = 0.1 L

Volume of a 0.200 M AgNO3 solution = 100 mL = 0.1 L

Initial temperature = 21.9 °C

Final temperature = 23.5 °C

Solid AgCl will be formed

<u>Step 2</u>: The balanced equation:

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

AgCl(s) + NaNO3(aq) → Na+(aq) + NO3-(aq) + AgCl(s)

<u>Step 3:</u> Define the formula

Pressure is constant.  → the heat evolved from the reaction is equivalent to the enthalpy of reaction.  

Q=m*c*ΔT

⇒ Q = the heat transfer (in joule)

⇒ m =the mass (in grams)

⇒ c= the heat capacity (J/g°C)

⇒ ΔT = Change in temperature = T2- T1

Step 4: Calculate heat

Let's vonsider the density the same as the density of water (1g/mL)

Mass = volume * density

Mass = 200 mL * 1g/mL

Mass = 200 grams

Q= m*c*ΔT

⇒ m = 200 grams

⇒ c = the heat capacity (let's consider the heat capacity of water) = 4.184 J/g°C

⇒ ΔT = 23.5 -21.9 = 1.6°C

Q = 200 * 4.184 * 1.6 = 1338 .9 J = 1.3 kJ

There is 1.3 kJ heat produced(released)

Therefore, we assumed no heat is absorbed by the calorimeter, no heat is exchanged between the  calorimeter and its surroundings, and the specific heat and mass of the solution are the same as those for  water (1g/mL and 4.184 J/g°C)

7 0
3 years ago
Mg(OH)2 + 2 HBr à MgBr2 + 2 H2O
AnnyKZ [126]

Explanation:

The balanced equation of the reaction is given as;

Mg(OH)2 (s) + 2 HBr (aq) → MgBr2 (aq) + 2 H2O (l)

1. How many grams of MgBr2 will be produced from 18.3 grams of HBr?

From the reaction;

2 mol of HBr produces 1 mol of  MgBr2

Converting to masses using;

Mass = Number of moles * Molar mass

Molar mass of HBr = 80.91 g/mol

Molar mass of MgBr2 = 184.113 g/mol

This means;

(2 * 80.91 = 161.82g) of HBr produces (1 * 184.113 = 184.113g) MgBr2

18.3g would produce x

161.82 = 184.113

18.3 = x

x = (184.113 * 18.3 ) / 161.82 = 20.8 g

2. How many moles of H2O will be produced from 18.3 grams of HBr?

Converting the mass to mol;

Number of moles = Mass / Molar mass = 18.3 / 80.91 = 0.226 mol

From the reaction;

2 mol of HBr produces 2 mol of H2O

0.226 mol would produce x

2 =2

0.226 = x

x = 0.226 * 2 / 2 = 0.226 mol

3. How many grams of Mg(OH)2 are needed to completely react with 18.3 grams of HBr?

From the reaction;

2 mol of HBr reacts with 1 mol of Mg(OH)2

18.3g of HBr =  0.226 mol

2 = 1

0.226 = x

x = 0.226 * 1 /2

x = 0.113 mol

5 0
3 years ago
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