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2 years ago

Both of the sulfur-oxygen double bonds in so2 are polar. in which direction should the polarity arrows point?

Chemistry

1 answer:

LiRa [457]2 years ago

8 0

Answer is: the polarity arrows should go from sulfur (S) to the oxygen (O).

The magnitude of the dipole depends on the difference in the electronegativies and has direction from positively charged atom to negatively charged atom.

Oxygen has greater electronegativity than sulfur, because of that oxygen (χ=3.5) is partially negative and sulfur (χ=2.5) is partially positive.

Electronegativity (χ) is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

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If you have a voltage of 120 volts entering your house and a light bulb with 0.625 amperes flowing through it, how many watts is

Luba_88 [7]

Answer:

75 watts

Explanation:

4 0

3 years ago

Read 2 more answers

If 125g of KClO3 is heated, what is the total mass of the products?

Andrej [43]

Given parameters:

Mass of KClO₃ = 125g

Unknown:

Total mass of the products = ?

When KClO₃ is heated, it thermally decomposes to KCl and O₂ according to the chemical equation below;

2KClO₃ → 2KCl + 3O₂

All chemical equations obeys the law of conservation of matter and with this regard, we know that the amount of reactants used is the same as that of the product.

The total mass of the products must give us 125g according to this law of conservation of matter.

Now to find the masses of each product,

Find the number of moles of the given reactant:

Number of moles = $\frac{mass}{molar mass}$

molar mass of KClO₃ = 39 + 35.5 + 3(16) = 122.5g/mol

So number of moles of KClO₃ = $\frac{125}{122.5}$ = 1.02moles

2. Now, using this number of moles, find the number of moles of the products using this value;

2 moles of KClO₃ produced 2 moles of KCl

1.02 moles of KClO₃ will also produce 1.02moles of KCl

2 moles of KClO₃ produced 3 moles of O₂

1.02 moles of KClO₃ will produce $\frac{1.02 x 3} {2}$ mole = 1.53 moles of O₂

3. Now find the masses of each product;

Mass = number of moles x molar mass

molar mass of KCl = 39 + 35.5 = 74.5g/mol

molar mass of O₂ = 16 x 2 = 32g/mol

Mass of KCl = 74.5 x 1.02 = 75.99g

Mass of O₂ = 32 x 1.53 = 48.96g

Total mass of products = mass of KCl + Mass of O₂ = 75.99g + 48.96g

= 124.95g

This value is approximately the same as that of mass of KClO₃

7 0

3 years ago

how many moles of a nonvolatile, nonelectrolyte solute are required to lower the freezing point of 1000 grams of water by 5.58

Ivan

Answer:

Explanation:

Using freezing point depression formula,

ΔTemp.f = Kf * b * i

Where,

ΔTemp.f = temp.f(pure solvent) - temp.f(solution)

b = molality

i = van't Hoff factor

Kf = cryoscopic constant

= 1.86°C/m for water

= (0 - (-5.58))/1.86

= 3.00 mol/kg

Assume 1 kg of water(solvent)

= (3.00 x 1)

= 3.00 mol.

5 0

3 years ago

What is the oxidation number of cl− in the hypochlorite ion clo−?

mariarad [96]

Answer : The oxidation number of chlorine (Cl) is, (+1)

Explanation :

Oxidation number : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

When the atoms are present in their elemental state then the oxidation number will be zero.

Rules for Oxidation Numbers :

The oxidation number of a free element is always zero.

The oxidation number of a monatomic ion equals the charge of the ion.

The oxidation number of Hydrogen (H) is +1, but it is -1 in when combined with less electronegative elements.

The oxidation number of oxygen (O) in compounds is usually -2, but it is -1 in peroxides.

The oxidation number of a Group 1 element in a compound is +1.

The oxidation number of a Group 2 element in a compound is +2.

The oxidation number of a Group 17 element in a binary compound is -1.

The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.

The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

The given compound is, $ClO^-$

Let the oxidation state of 'Cl' be, 'x'

$x+(-2)=-1\\\\x-2=-1\\\\x=+1$

Therefore, the oxidation number of chlorine (Cl) is, (+1)

5 0

3 years ago

Oxygen gas can be prepared by heating potassium chlorate according to the following equation:2KClO3(s)Arrow.gif2KCl(s) + 3O2(g)T

Eduardwww [97]

Answer:

Moles of potassium chlorate reacted = 0.2529 moles

The amount of oxygen gas collected will be 12.8675 g

Explanation:

(a)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 748 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (748 - 17.5) mmHg = 730.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 730.5 mmHg

V = Volume of the gas = 9.49 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$730.5mmHg\times 9.49L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{730.5\times 9.49}{62.3637\times 293}=0.3794mol$

According to the reaction shown below as:-

$2KClO_3(s)\rightarrow 2KCl(s) +3O_2(g)$

3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.

So,

0.3794 mol of oxygen gas are produced when $\frac{2}{3}\times 0.3794$ moles of potassium chlorate undergoes reaction.

<u>Moles of potassium chlorate reacted = 0.2529 moles</u>

(b)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 753 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (753 - 17.5) mmHg = 735.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 735.5 mmHg

V = Volume of the gas = 9.99 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$735.5mmHg\times 9.99L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{735.5\times 9.99}{62.3637\times 293}=0.40211mol$

Moles of Oxygen gas = 0.40211 moles

Molar mass of Oxygen gas = 32 g/mol

Putting values in above equation, we get:

$0.037mol=\frac{\text{Mass of Oxygen gas}}{2g/mol}\\\\\text{Mass of Oxygen gas}=(0.40211mol\times 32g/mol)=12.8675g$

<u>Hence, the amount of oxygen gas collected will be 12.8675 g</u>

5 0

2 years ago