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lutik1710 [3]
2 years ago
14

Please help

Chemistry
1 answer:
lora16 [44]2 years ago
6 0
Find the molar mass of CH3 and divide that by 45.0. That should give a whole number and then mult that whole number by CH3 to find molecular formula to get like ( incorrect ex: C2H6) which is mult by 2
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The atomic number of arsenic is 33 what is the electron configuration
rjkz [21]
It is really easy if you look up an electron configuration table (one that looks like a periodic table) and then just go down the rows left to right, top to bottom, and just stop when you get to where the element is on the table.

 1s^2 2s^2 2p^{6}  3s^2 3p^6 4s^2 3d^1^0 4p^3
8 0
3 years ago
What is the molar mass of an unknown gas with a density of 2.00 g/L at 1.00 atm and 25.0 °C?
soldier1979 [14.2K]

Answer:

Explanation:Explanation:

Your starting point here will be the ideal gas law equation

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

P

V

=

n

R

T

a

a

∣

∣

−−−−−−−−−−−−−−−

, where

P

- the pressure of the gas

V

- the volume it occupies

n

- the number of moles of gas

R

- the universal gas constant, usually given as

0.0821

atm

⋅

L

mol

⋅

K

T

- the absolute temperature of the gas

Now, you will have to manipulate this equation in order to find a relationship between the density of the gas,

ρ

, under those conditions for pressure and temperature, and its molar mass,

M

M

.

You know that the molar mass of a substance tells you the mass of exactly one mole of that substance. This means that for a given mass

m

of this gas, you can express its molar mass as the ratio between

m

and

n

, the number of moles it contains

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

M

M

=

m

n

a

a

∣

∣

−−−−−−−−−−−−−

(

1

)

Similarly, the density of the substance tells you the mass of exactly one unit of volume of that substance.

This means that for the mass

m

of this gas, you can express its density as the ratio between

m

and the volume it occupies

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

ρ

=

m

V

a

a

∣

∣

−−−−−−−−−−−

(

2

)

Plug equation

(

1

)

into the ideal gas law equation to get

P

V

=

m

M

M

⋅

R

T

Rearrange to get

P

V

⋅

M

M

=

m

⋅

R

T

P

⋅

M

M

=

m

V

⋅

R

T

M

M

=

m

V

⋅

R

T

P

Finally, use equation

(

2

)

to write

M

M

=

ρ

⋅

R

T

P

Convert the temperature of the gas from degrees Celsius to Kelvin then plug in your values to find

M

M

=

1.02

g

L

⋅

0.0821

atm

⋅

L

mol

⋅

K

⋅

(

273.15

+

37

)

K

0.990

atm

M

M

=

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

26.3 g mol

−

1

a

a

∣

∣

−−−−−−−−−−−−−−−−

I'll leave the answer rounded to three

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3 years ago
Which energy level has the most energy available to it?<br> 4<br> 3<br> 2<br> 1
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Tropic levels have most energy
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7 0
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N₂O(g) + 3 H₂(g) N₂H4(1) + H₂O(1) AH = -317 kJ/mol
docker41 [41]

Answer:

A

Explanation:

Recall that Δ<em>H</em> is the sum of the heats of formation of the products minus the heat of formation of the reactants multiplied by their respective coefficients. That is:


\displaystyle \Delta H^\circ_{rxn} = \sum \Delta H^\circ_{f} \left(\text{Products}\right) - \sum \Delta H^\circ_{f} \left(\text{Reactants}\right)

Therefore, from the chemical equation, we have that:


\displaystyle \begin{aligned} (-317\text{ kJ/mol}) = \left[\Delta H^\circ_f \text{ N$_2$H$_4$} +  \Delta H^\circ_f \text{ H$_2$O}  \right]   -\left[3 \Delta H^\circ_f \text{ H$_2$}+\Delta H^\circ_f \text{ N$_2$O}\right] \end{aligned}

Remember that the heat of formation of pure elements (e.g. H₂) are zero. Substitute in known values and solve for hydrazine:

\displaystyle \begin{aligned} (-317\text{ kJ/mol}) & = \left[ \Delta H^\circ _f \text{ N$_2$H$_4$} + (-285.8\text{ kJ/mol})\right] -\left[ 3(0) + (82.1\text{ kJ/mol})\right] \\ \\ \Delta H^\circ _f \text{ N$_2$H$_4$} & = (-317 + 285.8 + 82.1)\text{ kJ/mol} \\ \\ & = 50.9\text{ kJ/mol} \end{aligned}

In conclusion, our answer is A.

5 0
2 years ago
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