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Margaret [11]
4 years ago
5

What is the electron configuration for lithium

Chemistry
1 answer:
Natali5045456 [20]4 years ago
8 0
[He] 2s1 is the configuration for lithium.
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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
What is the volume of a 3.0 M solution of hydrochloric acid that contains 1.50 moles of solute?
charle [14.2K]
C - concentration: 3.0M
n - number of moles: 1.50 moles
V - volume: ???
____________
C = n/V
V = n/C
V = 1.5/3
V = 0,5 L = 500 mL

:)
4 0
4 years ago
The density of gold is 19.3 g/cm³. What is the mass of 11.3 cm³ of gold?
telo118 [61]

Answer:

218.09g

Explanation:

the formula for density is mass/volume

you have volume so rlly what you have is 19.3g/cm3=m/11.3cm3

so mass equals density x volume

and 19.3 x 11.3 = 218.09

4 0
3 years ago
Which of the following elements is the most reactive?
Zepler [3.9K]
Sodium. It reacts with most of every other element.
6 0
3 years ago
Read 2 more answers
How many formula units of iron(III) oxide can be produced from the reaction of 108 g of iron with 72.1 L of oxygen gas at STP?
Irina-Kira [14]

Answer:

5.81 x 10^23 formula units Fe2O3

Explanation:

5 0
3 years ago
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