Which of the following properties of an unknown substance would allow it to be classified as a pure substance ?

What term defines a mass which is exactly equal to 1/12 the mass of one carbon-12 atom

aalyn [17]

Answer:

Relative atomic mass and it is expressed in atomic mass unit (amu)

Explanation:

AMU:

Atomic mass unit is define as the 1/12 the mass of an atom of carbon-12.

C12 has six neutron and six protons in the nucleus.

This unit is used to express the masses of atoms. We know that masses of atoms are very small and we do not have any such type of balance that can measure very small quantity. So that is way we use this scale to measure small quantity. For example, according to this scale

relative atomic mass of hydrogen is 1.008 amu

relative atomic mass of oxygen is 15.999 amu

relative atomic mass of uranium is 238.0289 amu

relative atomic mass of chlorine is 35.453 amu

4 0

3 years ago

NaCl has a Delta. Hfus = 30. 2 kJ/mol. What is the mass of a sample of NaCl that needs 732. 6 kJ of heat to melt completely? Use

tia_tia [17]

The mass of NaCl sample has been 24.3 g. Thus, option A is correct.

The heat of fusion has been the amount of heat required to convert 1 mole of substance into solid to liquid state.

The heat required has been given as:

$Q=m\Delta H$

<h3>Computation for the mass of NaCl</h3>

The given solution has heat of fusion, $\Delta H_{fus}=30.2\rm kJ/mol$

The heat required to melt the sample has been, $Q=732.6\;\rm kJ/mol$

Substituting the values for the mass of NaCl

$\rm 732.6\;kJ/mol=Mass\;\times\;30.2\;kJ/mol\\\\
Mass=\dfrac{732.6}{30.2}\;g\\\\
 Mass=24.3\;g\\$

The mass of NaCl sample has been 24.3 g. Thus, option A is correct.

Learn more about heat of fusion, here:

brainly.com/question/87248

5 0

3 years ago

How many grams of F are in 12.56 g of SF6? h.

natali 33 [55]

Answer:

9.80 g

Explanation:

The molecular mass of the atoms mentioned in the question is as follows -

S = 32 g / mol

F = 19 g / mol

The molecular mass of the compound , SF₆ = 32 + ( 6 * 19 ) = 146 g / mol

The mass of 6 F = 6 * 19 = 114 g /mol .

The percentage of F in the compound =

mass of 6 F / total mass of the compound * 100

Hence ,

The percentage of F in the compound = 114 g /mol / 146 g / mol * 100

78.08 %

Hence , from the question ,

In 12.56 g of the compound ,

The grams of F = 0.7808 * 12.56 = 9.80 g

4 0

3 years ago

Why will a toy train remain still if you push on the engine with

netineya [11]

Explanation:

Because when two equal forces are applied from opposite directions, they "eliminate" each other.

The train would go right if a 3N force was only applied in the right direction, and it would go left if the same force was only applied to the left.

If a 3N force was applied to the right and a 2N force to the left, it would equal a 1N force to the right (3-2=1).

But there it's 3-3=0, so in practice the force is 0N. Therefore the train won't move.

6 0

3 years ago

Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic

Kisachek [45]

Answer:

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C) It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

$\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9$

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = $164.5\times 10^{-9} g$

$1 ng=10^{-9} g$

Volume of the sample = V = $15.3 cm^3$

Density of the lake water sample ,d= $1.00 g/cm^3$

Mass of sample = M = $d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g$

$ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75$

10.75 is the concentration of arsenic in the sample in parts per billion.

B)

Mass of arsenic in $1 cm^3$ of lake water = $\frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g$

Mass of arsenic in $0.710 km^3$ lake water be m.

$1 km^3=10^{15} cm^3$

Mass of arsenic in $0.710\times 10^{15} cm^3$ lake water :

$m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g$

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

$\frac{2.74}{41.90} days$

Then days required to remove 7,633.66 kg of arsenic from the lake water :

$=7,633.66\times \frac{2.74}{41.90} days=499.19 days$

1 year = 365 days

499.19 days = $\frac{499.19}{365} years = 1.367 years\approx 1.37 years$

It will take 1.37 years to remove all of the arsenic from the lake.

3 0

3 years ago