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1 year ago

A 60.2-ml sample of hg (density = 13.6 g/ml) contains how many atoms of hg?

1 answer:

6 0

Density is calculated using the following rule:
density = mass / volume
therefore:
mass = density * volume
mass of Hg = 13.6 * 60.2 = 818.72 grams

From the periodic table:
molar mass of Hg = 200.59 grams

number of moles = mass / molar mass
number of moles of Hg = 818.72 / 200.59 = 4.08 moles

each mole contains Avogadro's number of atoms.
Therefore,
number of atoms in the given sample = 4.08 * 6.022 * 10^23
= 2.456976 * 10^24 atoms

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Use average bond energies to calculate ΔHrxn for the following hydrogenation reaction: H2C=CH2(g)+H2(g)→H3C−CH3(g)

marissa [1.9K]

Answer:

The $\Delta H_{rxn}$ of the given reaction is -129.6 kJ

Explanation:

The given chemical reaction is as follows.

$H_{2}C=CH_{2}(g)+H_{2}(g)\rightarrow H_{3}C-CH_{3}(g)$

Enthalpy of each reactant and products are as follows.

$\Delta H_{C=C}\,=615.0\,kJ\,mol^{-1}$

$\Delta H _{H-H}\,=435.1\,kJ\,mol^{-1}$

$\Delta H _{C-C}\,=347.3\,kJ\,mol^{-1}$

$\Delta H _{C-H}\,=416.2\,kJ\,mol^{-1}$

In the given chemical reaction involved two C-H bonds in the reactant side and one C-C bond in the product side therefore, the enthalpy of formation will be the negative.

$\Delta H_{rxn}=-\Delta H_{C-C}-2\Delta H_{C-H}+\Delta H_{C=C}+\Delta H_{H-H}$

$=-347.4-2\times416.2+615.0+435.1$

$=-129.6 \,kJ$

Therefore, The $\Delta H_{rxn}$ of the given reaction is -129.6 kJ

4 0

1 year ago

Calculate the radius ratio for NaBr if the ionic radii of Na + and Br − are 102 pm and 196 pm , respectively. radius ratio: Base

Fudgin [204]

Answer : The expected coordination number of NaBr is, 6.

Explanation :

Cation-anion radius ratio : It is defined as the ratio of the ionic radius of the cation to the ionic radius of the anion in a cation-anion compound.

This is represented by,

$\frac{r_{cation}}{r_{anion}}$

When the radius ratio is greater than 0.155, then the compound will be stable.

Now we have to determine the radius ration for NaBr.

Given:

Radius of cation, $Na^+$ = 102 pm

Radius of cation, $Br^-$ = 196 pm

$\frac{r_{cation}}{r_{anion}}=\frac{102}{196}=0.520$

As per question, the radius of cation-anion ratio is between 0.414-0.732. So, the coordination number of NaBr will be, 6.

The relation between radius ratio and coordination number are shown below.

Therefore, the expected coordination number of NaBr is, 6.

8 0

2 years ago

What is the molar mass of al (clo2)3

frozen [14]

The molar mass is 229.33

4 0

1 year ago

Oxygen gas, generated by the reaction 2KClO3(s)---2KCl(s)+3O2(g), is collected over water at 27•C in 3.72L vassel at a total pre

Julli [10]

Answer:

moles = 0.093 moles

Explanation:

In this case, we know that this reaction is taking plave in a vessel that has a 730 torr of total pressure.

The total pressure is a value obtained by:

Pt = Pwater + PO2

We need to know the pressure of O2, because then, with stoichiometry, we can calculate the moles of KClO3

The pressure of oxygen is:

PO2 = 730 - 26 = 704 Torr

Now, this pressure is in Torr, and we need to convert it to Atm, so:

704 Torr / 760 Torr = 0.9263 atm

Now, let's use the ideal gas equation:

PV = nRT

With this expression, we will calculate the moles of O2, and then, the moles of KClO3:

n = PV/RT

R = 0.082 L atm /K mol

P = 0.9263 atm

V = 3.72 L

T = 27 + 273 = 300 K

Replacing the data:

n = 0.9263 * 3.72 / 300 * 0.082

n = 0.14 moles

Finally, by stoichiometry, we know that 2 moles of KClO3 produces 3 moles of O2, so:

moles of KClO3 = 0.14 * 2/3 = 0.093 moles of KClO3

6 0

1 year ago

2.00 liters of hydrogen, originally at 25.0 °C and 750.0 mm of mercury, are heated until a volume of 20.0 liters and a pressure

dedylja [7]

Answer: The new temperature is 10643 K

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 750.0 mm Hg = 0.98 atm (760mmHg=1atm)

$P_2$ = final pressure of gas = 3.50 atm

$V_1$ = initial volume of gas = 2.00 L

$V_2$ = final volume of gas = 20.0 L

$T_1$ = initial temperature of gas = $25.0^oC=273+25.0=298.0K$

$T_2$ = final temperature of gas = ?

Now put all the given values in the above equation, we get:

$\frac{0.98\times 2.00}{298.0K}=\frac{3.50\times 20.0}{T_2}$

$T_2=10643K$

Thus the new temperature is 10643 K

6 0

1 year ago