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4 years ago

The formula weight of potassium dichromate (k2cr2o7) is __________ amu.

Chemistry

1 answer:

Gre4nikov [31]4 years ago

4 0

The given compound is $K_{2}Cr_{2}O_{7}$

Each formula unit of $K_{2}Cr_{2}O_{7}$ has 2 mol Potassium (K) atoms, 2 mol Chromium (Cr) atoms and 7 mol Oxygen (O) atoms.

Atomic weight of potassium is 39.0983 amu

Atomic weight of Chromium is 51.9961 amu

Atomic weight of Oxygen is 15.9994 amu.

Therefore the formula weight of $K_{2}Cr_{2}O_{7} =$

$2 * 39.0983 amu + 2 * 51.9961 amu + 7 * 15.9994 amu$

= 294.1846 amu

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These four ELEMENTS were found in early earth and they make up all living things.

VladimirAG [237]

Answer: True

Air is a mixture of many things; it's not just Oxygen, but Hydrogen, Carbon, and Nitrogen as well are made up of air. Fun fact: Nitrogen is actually more abundant in air

6 0

3 years ago

Read 2 more answers

Suppose a new group 5A element is discovered. Photoelectron spectroscopy (PES) reveals that its last two peaks occur at 1370 kJ/

iren2701 [21]

Answer: The average valence electron energy (AVEE) of this element =

1014.2 KJ/ mol or 1.0142mJ/mol.

Explanation:

The average valence electron energy = (number of electrons in s subshell x Ionization energy of that subshell) + (number of electrons in p subshell x Ionization energy of that subshell) / total number of electrons in both subshells of the valence shells.

The 5A elements are non-metals like Nitrogen and Phosphorus with the metallic character increasing as you go down the group, So a new 5A element will have characteristics of its group with 5 valence electron in its outermost shell represented as ns2 np3

Therefore the average valence electron energy (AVEE) of this element will be calculated as

The average valence electron energy = (2 x 1370 kJ/mol + 3 x 777 kJ/mol.) / 5

2740+2331/ 5 =5071/5

=1014.2 KJ/ mol or 1.0142mJ/mol.

5 0

3 years ago

In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe

const2013 [10]

Explanation:

It is given that r = 0.283 nm. As 1 nm = $10^{-9} m$ .

Hence, 0.283 nm = $0.283 \times 10^{-9} m$

Formula for coulombic energy is as follows.

$U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}$

where, e = $1.6 \times 10^{-19}$ C

$\epsilon_{o}$ = $8.85 \times 10^{-12}$

$U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}$

= $1.423 \times 10^{-18} J$

As 1 eV = $1.6 \times 10^{-19} J$

So, 1 J = $\frac{1 eV}{1.6 \times 10^{-19}}$

Hence, U = $\frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}$

= 8.9 eV

Also, 1 J = $\frac{10^{-3} kJ}{6.022 \times 10^{23}mol}$

= $1.67 \times 10^{-27}$ kJ/mol

Therefore, U = $1.423 \times 10^{-18} J \times 1.67 \times 10^{-27}$ kJ/mol

= $2.37 \times 10^{-45} kJ/mol$

7 0

3 years ago

At what temperature (in C) will a sample of gas occupy 91.3 L if it occupies 45.0 L at 70.0°C? Assume constant pressure.)

sdas [7]

Solution is here,

for initial case,

temperature(T1)=70°C=70+ 273=343K

vloume( V1) =45 L

for final case,

temperature( T2)=?

volume(V2)= 91.3 L

at constant pressure,

V1/V2 = T1/T2

or, 45/91.3 = 343/ T2

or, T2= (343×91.3)/45

or, T2=695.9 K = (695.9-273)°C=422.9°C

5 0

3 years ago

Naturally occurring silicon has an atomic mass of 28.086 and consists of three isotopes. The major isotope is 28Si, natural abun

Elden [556K]

Answer:

29Si has a natural abundance of 4.68%.

30Si has a relative atomic mass of 29.99288 and a natural abundance of 3.09%.

Explanation:

The atomic mass of silicon is given by:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

Where:

Si: atomic mass of silicon (28.086)

Si²⁸: relative atomic mass of 28Si (27.97693)

A₁: natural abundance of 28Si (92.23%)

Si²⁹: relative atomic mass of 29Si (28.97649)

A₂: natural abundance of 29Si

Si³⁰: relative atomic mass of 30Si

A₃: natural abundance of 30Si

We also know that 30Si natural abundance is in the ratio of 0.6592 to that of 29Si.

We have to set up a system of three equations in three unknowns:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

A₃=0.6592×A₂

A₁+A₂+A₃=1

First, we find substitute the value of A₃ in the third equation and solv teh value of A₂:

A₁+A₂+0.6592×A₂=1

A₁+1.6592×A₂=1

1.6592×A₂=1-A₁

A₂= $\frac{1-A₁}{1.6592}=\frac{1-0.9223}{1.6592}$ =0.0468

Then, we find the value of A₃:

A₃=0.6592×A₂

A₃=0.6592×0.0468=0.0309

Finally, we find the value of Si³⁰ in the first equation:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

28.086=27.97693×0.9223+28.97649×0.0468+Si³⁰×0.0309

28.086=27.15922+Si³⁰×0.0309

28.086-27.15922=Si³⁰×0.0309

$\frac{0.92678}{0.0309}$ =Si³⁰

Si³⁰=29.99288

8 0

3 years ago