What happens to the humidity if the temperature drops and the amount of moisture in the air stays the same?

The element iridium exists in nature as two isotopes: 191Ir has a mass of 190.9606 u, and 193Ir has a mass of 192.9629 u. The av

nlexa [21]

Answer: The percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope be 'x'. So, fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope will be '1 - x'

For $_{77}^{191}\textrm{Ir}$ isotope:

Mass of $_{77}^{191}\textrm{Ir}$ isotope = 190.9606 amu

Fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope = x

For $_{77}^{193}\textrm{Ir}$ isotope:

Mass of $_{77}^{193}\textrm{Ir}$ isotope = 192.9629 amu

Fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope = 1 - x

Average atomic mass of iridium = 192.22 amu

Putting values in equation 1, we get:

$192.22=[(190.9606\times x)+(192.9629\times (1-x))]\\\\x=0.3710$

Percentage abundance of $_{77}^{191}\textrm{Ir}$ isotope = $0.3710\times 100=37.10\%$

Percentage abundance of $_{77}^{193}\textrm{Ir}$ isotope = $(1-0.3710)=0.6290\times 100=62.90\%$

Hence, the percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

8 0

2 years ago

The group number for alkali metals

kirill115 [55]

The alkali metals include: lithium, sodium, potassium, rubidium, cesium, and francium.

6 0

2 years ago

The combustion of propane (C3H8) produces CO2 and H2O: C3H8 (g) 5O2 (g) → 3CO2 (g) 4H2O (g) The reaction of 2.5 mol of O2 will p

olganol [36]

Answer: 2 moles of water.

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The combustion of propane is represented as :

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)$

According to stoichiometry:

5 moles of $O_2$ produces = 4 moles of $H_2O$

2.5 moles of $O_2$ will produce= $\frac{4}{5}\times 2.5=2$ moles of $H_2O$ .

Thus 2 moles of water are produced by 2.5 moles of oxygen.

3 0

2 years ago

A mole of X reacts at a constant pressure of 43.0 atm via the reaction X(g)+4Y(g)→2Z(g), ΔH∘=−75.0 kJ Before the reaction, the v

hoa [83]

Answer:

The total energy change, ΔE, in kilojoules = -61.93 kJ

Explanation:

Relationship between ΔH, ΔE and work done is given by first law of thermodynamics.

ΔE = ΔH - PΔV

Where,

ΔH = Change in enthalpy

ΔE = Change in internal energy

PΔV = Work done

Given that,

ΔH = -75.0 kJ = -75000 J

P = 43.0 atm

ΔV = Final volume - initial volume

= (2.00 - 5.00) = -3.00 L

PΔV = 43 × (-3.00) = -129 L atm

1 L atm = 101.325 J

-129 L atm = 129 × 101.325 = -13071 J

So ,

ΔE = ΔH - PΔV

= (-75000 J) - ( -13071 J)

= -75000 J + 13071 J

= -61929 J

Total energy change, ΔE = -61.929 kJ

3 0

2 years ago

Need help with these two pls!!!!

Allushta [10]

Answer:

the first questions answer is number two a tool for starting afire invented in southeast asia

Explanation:

8 0

2 years ago