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jasenka [17]
3 years ago
5

Cuántos gramos de cloruro de magnesio se pueden obtener si se hacen reaccionar 354,9g de cloruro de titanio (IV) y 285g de magne

sio puro, Recuerde que como productos obtiene titanio y cloruro de magnesio.
a. escriba la ecuación balanceada

b. determine el reactivo limite

c. La cantidad de reactivo en exceso
Chemistry
1 answer:
DENIUS [597]3 years ago
8 0

Answer:

sorry I dont speak Spanish

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Gold forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimeter for a silver-gold
Mama L [17]

<u>Answer:</u> The number of gold atoms per cubic centimeters in the given alloy is 1.83\times 10^{22}

<u>Explanation:</u>

To calculate the number of gold atoms per cubic centimeters for te given silver-gold alloy, we use the equation:

N_{Au}=\frac{N_AC_{Au}}{(\frac{C_{Au}M_{Au}}{\rho_{Au}})+(\frac{M_{Au}(100-C_{Au})}{\rho_{Ag}})}

where,

N_{Au} = number of gold atoms per cubic centimeters

N_A = Avogadro's number = 6.022\times 10^{23}atoms/mol

C_{Au} = Mass percent of gold in the alloy = 42 %

\rho_{Au} = Density of pure gold = 19.32g/cm^3

\rho_{Ag} = Density of pure silver = 10.49g/cm^3

M_{Au = molar mass of gold = 196.97 g/mol

Putting values in above equation, we get:

N_{Au}=\frac{(6.022\times 10^{23}atoms/mol)\times 48\%}{(\frac{48\%\times 196.97g/mol}{19.32g/cm^3})+(\frac{196.97g/mol\times 58\%}{10.49g/cm^3})}\\\\N_{Au}=1.83\times 10^{22}atoms/cm^3

Hence, the number of gold atoms per cubic centimeters in the given alloy is 1.83\times 10^{22}

5 0
3 years ago
Is brass a solution:
Ivahew [28]

Answer:

Brass is an alloy, and either a "solid solution".

Alloys in general may be solid solutions or they simply be mixtures

Explanation:

<h2>Hope it Helps you!! </h2>
3 0
3 years ago
Please help thank you
kondaur [170]

Answer:

C. The hand shears because their shorter handles transfer force more quickly to the cutting blade.

Explanation:

6 0
3 years ago
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Why are liquids often dense?
Sophie [7]
The answer: The answer is B
8 0
3 years ago
An element has two naturally occurring isotopes with atomic masses of 112.90 amu and 114.90 amu. The relative abundances of thes
Musya8 [376]

Answer:

C. 114.8 u

Explanation:

The atomic mass of X is the <em>weighted average</em> of the atomic masses of its isotopes.

We multiply the atomic mass of each isotope by a number representing its relative importance (i.e., its percent of the total).

Set up a table for easy calculation

0.0429 × 112.90 u =     4.843 u

0.9571  × 114.90 u = <u>109.97    u </u>

                 TOTAL = 114.8       u

7 0
3 years ago
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