What is the molar mass of nitrogen?

Write an electron configuration for an atom of aluminum-27 in an excited state

andrew11 [14]

Ground state of Al is 1s²2s²2p⁶3s²3p¹.
Exited state of Al is 1s²2s²2p⁶3s¹3p².

8 0

3 years ago

2.3 Zinc has five naturally occurring isotopes: 48.63% of 64 Zn with an atomic weight of 63.929 amu; 27.90% of 66Zn with an atom

lakkis [162]

Answer: The average atomic mass of element Zinc is 65.40 amu.

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)

For $_{30}^{64}\textrm{Zn}$ isotope:

Mass of $_{30}^{64}\textrm{Zn}$ isotope = 63.929 amu

Percentage abundance of $_{30}^{64}\textrm{Zn}$ isotope = 48.63 %

Fractional abundance of $_{30}^{64}\textrm{Zn}$ isotope = 0.4863

For $_{30}^{66}\textrm{Zn}$ isotope:

Mass of $_{30}^{66}\textrm{Zn}$ isotope = 65.926 amu

Percentage abundance of $_{30}^{66}\textrm{Zn}$ isotope = 27.90 %

Fractional abundance of $_{30}^{66}\textrm{Zn}$ isotope = 0.2790

For $_{30}^{67}\textrm{Zn}$ isotope:

Mass of $_{30}^{67}\textrm{Zn}$ isotope = 66.927 amu

Percentage abundance of $_{30}^{67}\textrm{Zn}$ isotope = 4.10 %

Fractional abundance of $_{30}^{67}\textrm{Zn}$ isotope = 0.0410

For $_{30}^{68}\textrm{Zn}$ isotope:

Mass of $_{30}^{68}\textrm{Zn}$ isotope = 67.925 amu

Percentage abundance of $_{30}^{68}\textrm{Zn}$ isotope = 18.75 %

Fractional abundance of $_{30}^{68}\textrm{Zn}$ isotope = 0.1875

For $_{30}^{70}\textrm{Zn}$ isotope:

Mass of $_{30}^{70}\textrm{Zn}$ isotope = 69.925 amu

Percentage abundance of $_{30}^{70}\textrm{Zn}$ isotope = 0.62 %

Fractional abundance of $_{30}^{70}\textrm{Zn}$ isotope = 0.0062

Putting values in equation 1, we get:

$\text{Average atomic mass of Zinc}=[(63.929\times 0.4863)+(65.926\times 0.2790)+(66.927\times 0.0410)+(67.925\times 0.1875)+(69.925\times 0.0062)]$

$\text{Average atomic mass of Zinc}=65.40amu$

Hence, the average atomic mass of element Zinc is 65.40 amu.

7 0

3 years ago

What is the molarity of a solution containing 7.47 moles of solute in 4.31 liters of solvent?

scoray [572]

Answer:

1.73 M

Explanation:

Molarity is moles per liter, so we need to divide 7.47 moles by 4.31 liters to get the molarity of the solution.

7.47/4.31 ≈ 1.73 M

5 0

3 years ago

A volume of 129 mL of hydrogen is collected over water. The water level in the collecting vessel is the same as the outside leve

mixas84 [53]

Explanation:

As it is given that water level is same as outside which means that theoretically, P = 756.0 torr.

So, using ideal gas equation we will calculate the number of moles as follows.

PV = nRT

or, n = $\frac{PV}{RT}$

= $\frac{\frac{756}{760}atm \times 0.129 L}{0.0821 Latm/mol K \times 298 K}$

= 0.0052 mol

Also, No. of moles = $\frac{mass}{\text{molar mass}}$

0.0052 mol = $\frac{mass}{2 g/mol}$

mass = 0.0104 g

As some of the water over which the hydrogen gas has been collected is present in the form of water vapor. Therefore, at $25^{o}C$

$P_{\text{water vapor}}$ = 24 mm Hg

= $\frac{24}{760}$ atm

= 0.03158 atm

Now, P = $\frac{756}{760} - 0.03158$

= 0.963 atm

Hence, n = $\frac{0.963 atm \times 0.129 L}{0.0821 L atm/mol K \times 298 K}$

= 0.0056 mol

So, mass of $H_{2}$ = 0.0056 mol × 2

= 0.01013 g (actual yield)

Therefore, calculate the percentage yield as follows.

Percent yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$

= $\frac{0.01013 g}{0.0104 g} \times 100$

= 97.49%

Thus, we can conclude that the percent yield of hydrogen for the given reaction is 97.49%.

3 0

3 years ago

Help out please?????

Alexxandr [17]

Answer:

A non- metal

Explanation:

Hope u got ur answer

5 0

3 years ago

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