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2 years ago

The molar mass of an element is the mass of one

Chemistry

1 answer:

allochka39001 [22]2 years ago

7 0

Answer:

The molar mass of an element is the mass of one mole of atoms of the element.

Explanation:

The molar mass of an element is its atomic mass, i.e. the mass in grams of one mole of atoms of the element.

Remember 1 mol is approximately 6.022 × 10²³.

So, 1 mol of atoms is 6.022 × 10²³ atoms.

The molar mass is an average: it is the weighted average mass of the natural isotopes of the element, taking into account their relative abundance.

For example, the molar mass or atomic mass of carbon is 12,0107 g/mol, instead of 12.0000, becasue carbon exists in several forms (isotopes), and so the weighted average is not a whole number.

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How many grams of aluminum chloride are needed to react completely with 1.084g lithium sulfide?

Oksi-84 [34.3K]

Molar mass :

Li₂S = 45.947 g/mol

AlCl₃ = 133.34 g/mol

3 Li₂S + 2 AlCl₃ = 6 LiCl + Al₂S₃

3 * 45.947 g Li₂S ----------> 2 * 133.34 g AlCl₃
1.084 g Li₂S ----------------> ?

Mass Li₂S = 1.084 * 2 * 133.34 / 3 * 45.947

Mass Li₂S = 289.08112 / 137.841

Mass Li₂S = 2.0972 g

hope this helps!

5 0

2 years ago

The temperature of 6.24 L of a gas is increased from 25.0°C to 55.0°C at constant pressure. The new volume of the gas is Questio

Sphinxa [80]

Answer:

Heating this gas to 55 °C will raise its volume to 6.87 liters.

Assumption: this gas is ideal.

Explanation:

By Charles's Law, under constant pressure the volume $V$ of an ideal gas is proportional to its absolute temperature $T$ (the one in degrees Kelvins.)

Alternatively, consider the ideal gas law:

$\displaystyle V = \frac{n \cdot R}{P}\cdot T$ .

$n$ is the number of moles of particles in this gas. $n$ should be constant as long as the container does not leak.
$R$ is the ideal gas constant.
$P$ is the pressure on the gas. The question states that the pressure on this gas is constant.

Therefore the volume of the gas is proportional to its absolute temperature.

Either way,

$V\propto T$ .

$\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1}$ .

For the gas in this question:

Initial volume: $V_1 = \rm 6.24\; L$ .

Convert the two temperatures to degrees Kelvins:

Initial temperature: $T_1 = \rm 25.0\;\textdegree{C} = (25.0 + {\rm 273.15})\; K = 298.15\;K$ .
Final temperature: $T_1 = \rm 55.0\;\textdegree{C} = (55.0 + {\rm 273.15})\; K = 328.15\;K$ .

Apply Charles's Law:

$\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1} = \rm 6.24\;L \times \frac{328.15\; K}{298.15\;K} = 6.87\;L$ .

7 0

3 years ago

Explain why the standard enthalpy of formation () for Cl2 (g) is zero, but its standard entropy is larger than zero.

Galina-37 [17]

The standard enthalpy of formation for chlorine is zero but the standard entropy is larger than 0 because it is the elemental state of chlorine.

The standard enthalpy of formation for chlorine is zero because cl2 is the elemental state of chlorine and it does not require any energy for the formation of the standard state of chlorine.

The entropy of any system cannot be negative. It can only be positive or zero.

The entropy of a system will become zero only at a absolute zero temperature.

That's why the entropy of chlorine in elemental state is more than zero because absolutely zero temperature can't be obtained.

To know more about entropy, visit,

brainly.com/question/6364271

#SPJ4

7 0

11 months ago

State the period and group to which aluminum belongs to in the periodic table

FinnZ [79.3K]

Explanation:

aluminium belongs to group 13 and period 3

3 0

3 years ago

How many grams of aluminum oxide will you need to start with in order to make 6500 g of aluminum hydroxide?

kherson [118]

Answer:

Explanation:

2Al(s) + 3 2 O2(g) → Al2O3(s) And given the stoichiometry ...and EXCESS dioxygen gas...we would get 6.25⋅ mol of alumina. the which represents a mass... ...6.25 ⋅ mol ×101.96 ⋅ g ⋅ mol−1 molar mass of alumina ≡ 637.25 ⋅ g.

3 0

2 years ago

Read 2 more answers