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Consider the balanced equation below.

wolverine [178]

Answer:

a) H2S:SO2 = 2:2 and O2:H2O = 3:2

Explanation:

2H2S + 3O2 → 2SO2 + 2H2O

⇒ H2S:SO2 = 2 mol H2S / 2 mol SO2 ≡ 2 : 2

⇒ O2:H2O = 3 mol O2 / 2 mol H2O ≡ 3 : 2

8 0

3 years ago

If 185 g of KBr are dissolved in 1.2 kg of water, what would be the expected change in boiling point? The boiling point constant

Fudgin [204]

Answer:

ΔTb = 0.66 C

Explanation:

Given

Mass of KBr = 185 g

Mass of water = 1.2 kg

Kb = 0.51 C/m

Explanation:

The change in boiling point (ΔTb) is given by the product of molality (m) of the solution and the boiling point constant (Kb)

$\Delta T_{b}= K_{b}* m$

$Molality = \frac{moles\ KBr}{Kg\ water} \\\\moles KBr = \frac{mass\ KBr}{Mol.wt\ KBr} = \frac{185}{119} = 1.555\\\\Molality (m) = \frac{1.555 }{1.2} =1.296 m\\$

[tex]\Delta T_{b}= 0.51 C/m * 1.296 m = 0.66 C[\tex]

6 0

4 years ago

Read 2 more answers

Which of the following solids would possess the lowest melting point?

saul85 [17]

The melting points of the compounds from the choices are listed below:

a. C25H52 -> +47 to +56 degrees celsius

b. Hg -> -38.83 degrees celsius

c. CaCl2 -> +772 degrees celsius

d. CuO -> +1326 degrees celsius

It is evident from the listed melting points above that mercury (Hg) has the lowest melting point. This is why mercury is liquid at room temperature.

4 0

3 years ago

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hydrogen and oxygene react to form water l. How many molecules of O2 are required to produce 0.6 of H20

user100 [1]

Answer:

It is required 0.3 molecules of O2 to produce 0.6 molecules of H2O

Explanation:

The water´s formation reaction is: 2H2 + O2 --> 2H2O

This reaction shows the molecular O2:H2O ratio: 1:2.

Then, if we want to know how many molecules of O2 are required to produce 0.6 of H20, it is necessary calculate as it showed next:

0.6 molecules H2O * (1 molecula O2 / 2 molecules H2O)= 0.3 molecule O2

8 0

3 years ago

Menthol is a flavoring agent extracted from peppermint oil. It contains C, H, and O. In one combustion analysis, 10.00 mg of the

RideAnS [48]

Answer:

C₁₀H₂₀O

Explanation:

The molecular formula must be $C_{x}H_yOz$ . The combustion reaction will occur between the fuel and oxygen gas:

$C_{x}H_yOz$ + O₂ → CO₂ + H₂O

For Lavoisier law, the mass of the reactants must be equal to the mass of the products (mass conservation):

10.0 + mO₂ = 28.16 + 11.53

mO₂ = 29.69 mg

Supposing that all the oxygen and the menthol were consumed, let's calculate the number of moles of the compounds, knowing, for the Periodic Table, that:

MC = 12 g/mol, MO = 16 g/mol, MH = 1 g/mol

MCO₂ = 12 + 2x16 = 44 g/mol

MH₂O = 2x1 + 16 = 18 g/mol

MO₂ = 2x16 = 32 g/mol

n = mass (g)/molar mass

nCO₂ = 0.02816/44 = 6.4x10⁻⁴ mol

nH₂O = 0.01153/18 = 6.4x10⁻⁴ mol

nO₂ = 0.02969/32 = 9.3x10⁻⁴ mol

The molar number is proportional in the molecule, so, in CO₂, the number of C is 6.4x10⁻⁴ mol, and of O is 1.28x10⁻³. All the carbon of the methol will be in CO₂, and all the H will be in H₂O. The number of moles of O will be the difference of moles in H₂O and the O₂, then:

nC = 6.4x10⁻⁴ mol

nH = 2x6.4x10⁻⁴ = 1.28x10⁻³ mol

nO = (2x6.4x10⁻⁴ + 6.4x10⁻⁴) - (2x9.3x10⁻⁴) = 6.0x10⁻⁵

The empirical formula is the molecule formula with the small subscripts numbers, which represent the number of moles of the atoms in the molecule. So, let's divide all the number of moles for the small on 6.0x10⁻⁵.

nC = (6.4x10⁻⁴)/(6.0x10⁻⁵) = 10

nH = (1.28x10⁻³)/(6.0x10⁻⁵) = 20

nO = (6.0x10⁻⁵)/(6.0x10⁻⁵) = 1

So, the empirical formula of methol is C₁₀H₂₀O.

3 0

3 years ago