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3 years ago

A sample of gold (Au) has a mass of 45.39 g.

1 answer:

8 0

1. What is the molar mass of gold?
Molar mass is a unit that expresses the mass of a molecule per one mol. The molar mass can be obtained by adding the neutron with the proton of the atoms. Gold has atomic number 79 so the proton is 79. The number of the neutron is 118. Then the molar mass would be: 79 + 118 = 197 g/mol

2. Calculate the number of moles of gold (Au) in the sample. Show your work.
In this question, you are given the mass of the gold and asked for how many moles the sample has. To find the number of moles you just need to divide the weight by the molar mass.
For 45.39 grams of gold, the number of moles would be:
45.39 / (197g/mol)= 0.23 moles

3. Calculate the number of atoms of gold (Au) in the sample. Show your work.Moles is unit of a number of molecules but 1 mol doesn't represent 1 molecule. The number of atoms can be obtained by multiplying the number of moles with Avogadro number. The calculation would be:
0.23 moles * (6.023 * 10^23 molecules/mol)= 1.387 * 10^23 molecules

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a 2.7 L of N2 is collected at 121kpa and 288 K . if the pressure increases to 202 kpa and the temperature rises to 303 K , what

jok3333 [9.3K]

Answer:

The gas will occupy a volume of 1.702 liters.

Explanation:

Let suppose that the gas behaves ideally. The equation of state for ideal gas is:

$P\cdot V = n\cdot R_{u}\cdot T$ (1)

Where:

$P$ - Pressure, measured in kilopascals.

$V$ - Volume, measured in liters.

$n$ - Molar quantity, measured in moles.

$T$ - Temperature, measured in Kelvin.

$R_{u}$ - Ideal gas constant, measured in kilopascal-liters per mole-Kelvin.

We can simplify the equation by constructing the following relationship:

$\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}}$ (2)

Where:

$P_{1}$ , $P_{2}$ - Initial and final pressure, measured in kilopascals.

$V_{1}$ , $V_{2}$ - Initial and final volume, measured in liters.

$T_{1}$ , $T_{2}$ - Initial and final temperature, measured in Kelvin.

If we know that $P_{1} = 121\,kPa$ , $P_{2} = 202\,kPa$ , $V_{1} = 2.7\,L$ , $T_{1} = 288\,K$ and $T_{2} = 303\,K$ , the final volume of the gas is:

$V_{2} = \left(\frac{T_{2}}{T_{1}} \right)\cdot \left(\frac{P_{1}}{P_{2}} \right)\cdot V_{1}$

$V_{2} = 1.702\,L$

The gas will occupy a volume of 1.702 liters.

6 0

3 years ago

Help me please help...

Mashcka [7]

Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases. -google

its probably 40

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2 years ago

Which of the following is accurate in describing the placement and classification of iodine?

fenix001 [56]

You can answer this question by only searching the element in the periodic table.

The atomic number of iodine, I, is 53. It is placed in the column 17 (this is the Group) and row 5 (this is the Period).

The conclusion is that the iodine is located in Period 5, Group 17, and is classified as a nonmetal.

4 0

2 years ago

Is o-toluic acid soluble in ether, NaOH?, Is Napthalene soluble in NaOH?

natima [27]

Yes, o-toluic acid is soluble in ether as ether is slightly polar and it is soluble in NaOH because it is likely to form soluble compounds with it.

Naphthalene is insoluble in NaOH.

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3 years ago

Give the name of reaction in the preparation of alkene from alcohol compounds.

pickupchik [31]

Dehydration to alkenes!

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3 years ago