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DanielleElmas [232]

3 years ago

9

How many N2 molecules would exist in 500g of water at 20C?

1 answer:

tankabanditka [31]3 years ago

6 0

We have to know the number of molecules present in 500 g of water.

The answer is: zero molecules present in 500g of water.

Water molecule is H₂O. In water molecule, nitrogen molecule can not be present. Molecular mass of water is 18 g. MOlecular mass of nitrogen molecule is 28 g.

500 g water contains 500/18 number of moles= 27.77 moles of water molecules which contains 27.77 X 6.023 X 10²³ number of water molecules.

There is no existence of N₂ molecule.

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6. Using the rules of significant figures, calculate the following:

Burka [1]

Answer:

A. 3

B . 2

C. 2

D. 2

E. 4 significant figures

3 0

3 years ago

PLZ HELP For the reaction: 2NO2(g) → N2O4(l),

Furkat [3]

Answer: $\Delta H_{rxn}=-20kJ/mol-(+66kJ/mol)$

Explanation:

Heat of reaction or enthalpy change is the energy released or absorbed during the course of the reaction.

It is calculated by subtracting the enthalpy of reactants from the enthalpy of products.

$\Delta H=H_{products}-H_{reactants}$

$\Delta H$ = enthalpy change = ?

$H_{products}$ = enthalpy of products

$H_{reactants}$ = enthalpy of reactants

For the given reaction :

$2NO_2(g)\rightarrow N_2O_4(l)$

$\Delta H=H_{N_2O_4}-2\times H_{NO_2}$

$\Delta H=-20kJ/mol-(+66kJ/mol)$

6 0

3 years ago

Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t

Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

$(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)$

Now we have to calculate the amount of heat released or absorbed in both processes.

<u>For process 1 :</u>

$Q_1=m\times \Delta H_{fusion}$

where,

$Q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

$\Delta H_{fusion}$ = enthalpy change for fusion = $3.35\times 10^5J/Kg$

Now put all the given values in $Q_1$ , we get:

$Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J$

<u>For process 2 :</u>

$Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})$

where,

$Q_2$ = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

$c_{p,l}$ = specific heat of liquid water = $4186J/Kg^oC$

$T_1$ = initial temperature = $0^oC$

$T_2$ = final temperature = $100^oC$

Now put all the given values in $Q_2$ , we get:

$Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC$

$Q_2=4.186\times 10^5J$

From this we conclude that, $Q_1$ that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

5 0

3 years ago

Express the concentration of a 0.0320 M aqueous solution of fluoride, F−, in mass percentage and in parts per million (ppm). Ass

denis23 [38]

Answer:

607 ppm

Explanation:

In this case we can start with the <u>ppm formula</u>:

$ppm=\frac{mg~of~solute}{Litters~of~solution}$

If we have a solution of <u>0.0320 M</u>, we can say that in 1 L we have 0.032 mol of $F^-$ , because the molarity formula is:

$M=\frac{mol}{L}$

In other words:

$0.0320~M=\frac{mol}{1~L}$

$mol=0.032~M*1~L=0.032~mol$

$1~L~of~Solution=0.0320~mol~of~solute$

If we use the <u>atomic mass</u> of $F$ (19 g/mol) we can convert from mol to g:

$0.0320~mol~F^-\frac{19~g~F^-}{1~mol~F^-}~=~0.607~g$

Now we can <u>convert from g to mg</u> (1 g= 1000 mg), so:

$0.607~g\frac{1000~mg}{1~g}=607~mg$

Finally we can <u>divide by 1 L</u> to find the ppm:

$ppm=\frac{607~mg}{1~L}=~607~ppm$

<u>We will have a concentration of 607 ppm.</u>

I hope it helps!

4 0

4 years ago

When making calculations, you should rely on the precision of your measured data.

ELEN [110]

True

Im happy i could help you today

have a great rest of ur week :)

3 0

3 years ago