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Ber [7]
3 years ago
14

The number of atoms in 2 moles of aluminium is if the 1mole is 6.022*10 the power of 23 or 10^23

Chemistry
1 answer:
patriot [66]3 years ago
3 0

Answer:

1.204 × 10²⁴ atoms

Explanation:

According to this question, one mole of aluminum (Al) atom contains 6.02 × 10²³ atoms.

If two moles of aluminum are given, this means that there will be 2 × 6.02 × 10²³ atoms of aluminum

2 × 6.02 × 10²3

= 12.04 × 10²³

= 1.204 × 10²⁴ atoms.

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Oil doesn't dissolve in water, which could be your answer.
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A sample of a compound that contains only the elements C, H, and N is completely burned in O₂ to produce 44.0 g of CO₂, 45.0 g o
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Answer:

CH₅N

Explanation:

In the combustion, all of the C in the compound was used to produce CO₂ in a 1:1 ratio. Thus, the moles of CO₂ (MW 44.01 g/mol) produced equals the moles of C in the compound:

(44.0 g)(mol/44.01g) = 0.99977 mol CO₂ = 0.99977... mol C

Similarly, all of the H in the compound was used to produce H₂O in a ratio of 2H:1H₂O. The moles of H₂O (MW 18.02 g/mol) produced was:

(45.0 g)(mol/18.02g) = 2.497...mol H₂O

Moles of H is found using the molar ratio of 2H:1H₂O:

(2.497...mol H₂O)(2H/1H₂O) = 4.994...mol H

The ratio of H to C in the compound is:

(4.994...mol H)/(0.99977... mol C) = 5 H:C

Some NO₂ was produced from the N in the compound. Assuming a 1:1 ratio of C:N, the simplest empirical formula is: CH₅N.

3 0
3 years ago
How to writhe the balanced chemical equation
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Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 1.30 MJ o
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Answer:

97 000 g Na

Explanation:

The absortion (or liberation) of energy in form of heat is expressed by:

q=m*Cp*ΔT

The information we have:

q=1.30MJ= 1.30*10^6 J

ΔT = 10.0°C = 10.0 K (ΔT is the same in °C than in K)

Cp=30.8 J/(K mol Na)

If you notice, the Cp in the question is in relation with mol of Na. Before using the q equation, we can find the Cp in relation to the grams of Na.

To do so, we use the molar mass of Na= 22.99g/mol

Cp= \frac{30.8J}{K*mol Na}*\frac{1 mol Na}{22.99 g Na}=1.34\frac{J}{K*g Na}

Now, we are able to solve for m:

m=\frac{1.30*10^6 J}{1.34\frac{J}{K*g Na} *10.0K}=\frac{1.30*10^6J}{13.4\frac{J}{g Na} }  = 9.70*10^4 g Na=97 Kg Na=97 000 g Na

7 0
3 years ago
Read 2 more answers
Copper has a density of 8.96 g/cm3. If 75.0 g of copper is added to 50.0 mL of water in a graduated cylinder, to what volume rea
Tcecarenko [31]

Answer:

The answer to your question is    Final volume = 58.37 ml

Explanation:

Data

density = 8.96 g/cm³

mass = 75 g

volume of water = 50 ml

Process

1.- Calculate the volume of copper

  Density = mass / volume

Solve for volume

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Substitution

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Simplification

  Volume = 8.37cm³    or 8.37 cm³

2.- Calculate the new volume of water in the graduated cylinder

  Final volume = 50 + 8.37

  Final volume = 58.37 ml

3 0
3 years ago
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