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1 year ago

Is each oxide basic, acidic, or amphoteric in water: (a) SeO₂ ;

Chemistry

1 answer:

SpyIntel [72]1 year ago

3 0

Since SeO2 dissolves in water to produce selenium acid, it is regarded as an acidic oxide.

The chemical compound having the formula SeO2 is known as selenium dioxide. One of the forms of selenium that is most frequently seen is this colorless solid. 350 °C (662 °F; 623 K). The chemical compound having the formula SeO2 is known as selenium dioxide. One of the forms of selenium that is most frequently seen is this colorless solid. It serves as a catalyst, reagent, oxidizing agent, and antioxidant in lubricating oils. OSHA regulates selenium oxide; thus, it is listed on the Hazardous Substances List. It can be produced chemically by dehydrating selenous acid or chemically by burning selenium in oxygen, nitrogen dioxide, or hydrogen peroxide.

Learn more about Selenium dioxide here-

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the sum of the numbers of protons and neutrons present in the nucleus of an atom.

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Why aluminum in 1820 had more value than gold

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2 years ago

What is the mass of 4.75 mol H2SO4

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3 years ago

If you have a 150-gram sample of CrO3: • How many moles of CrO3 do you have? • How many oxygen atoms do you have? • How many gra

docker41 [41]

Answer :

(a) The moles of $CrO_3$ is, 1.5 moles.

(b) The number of oxygen atoms are, $27.099\times 10^{23}$

(c) The mass of oxygen is, 72 grams.

Explanation : Given,

Mass of $CrO_3$ = 150 g

Molar mass of $CrO_3$ = 100 g/mole

Molar mass of oxygen = 16 g/mole

(a) First we have to calculate the moles of $CrO_3$ .

$\text{Moles of }CrO_3=\frac{\text{Mass of }CrO_3}{\text{Molar mass of }CrO_3}=\frac{150g}{100g/mole}=1.5moles$

The moles of $CrO_3$ is, 1.5 moles.

(b) Now we have to calculate the number of oxygen atoms.

In $CrO_3$ , there are 1 atom of chromium and 3 atoms of oxygen.

According to the mole concept,

1 mole of $CrO_3$ contains $3\times 6.022\times 10^{23}$ number of oxygen atoms.

So, 1.5 mole of $CrO_3$ contains $1.5\times 3\times 6.022\times 10^{23}=27.099\times 10^{23}$ number of oxygen atoms.

The number of oxygen atoms are, $27.099\times 10^{23}$

(c) Now we have to calculate the mass of oxygen.

As, 1 mole of $CrO_3$ contains 3 moles of oxygen

So, 1.5 mole of $CrO_3$ contains $1.5\times 3=4.5$ moles of oxygen

$\text{Mass of oxygen}=\text{Moles of oxygen}\times \text{Molar mass of oxygen}$

$\text{Mass of oxygen}=(4.5mole)\times (16g/mole)=72g$

The mass of oxygen is, 72 grams.

3 0

3 years ago

The normal boiling point of bromine is 58.8°C, and its enthalpy of vaporization is 30.91 kJ/mol. What is the approximate vapor p

saul85 [17]

Answer : The vapor pressure of bromine at $10.0^oC$ is 0.1448 atm.

Explanation :

The Clausius- Clapeyron equation is :

$\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = vapor pressure of bromine at $10.0^oC$ = ?

$P_2$ = vapor pressure of propane at normal boiling point = 1 atm

$T_1$ = temperature of propane = $10.0^oC=273+10.0=283.0K$

$T_2$ = normal boiling point of bromine = $58.8^oC=273+58.8=331.8K$

$\Delta H_{vap}$ = heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

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

$\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})$

$P_1=0.1448atm$

Hence, the vapor pressure of bromine at $10.0^oC$ is 0.1448 atm.

4 0

3 years ago