A mystery element contains 11 electrons. Which group of the periodic table will it belong to?

What volume of 0.210 M sulfuric acid is required to completely react with 2.14 g aluminium hydroxide?

Galina-37 [17]

3 H2SO4 + 2 Al(OH)3 → Al2(SO4)3 + 6 H2O

(2.14 g Al(OH)3) / (78.0036 g Al(OH)3/mol) x (3 mol H2SO4 / 2 mol Al(OH)3) / (0.210 mol/L H2SO4) =
0.19596 L = 196 mL H2SO4

3 0

1 year ago

What is the difference in height between the top surface of the glycerin and the top surface of the alcohol? Suppose that the de

Snezhnost [94]

Here is the full question

Glycerin is poured into an open U-shaped tube until the height in both sides is 20 cm. Ethyl alcohol is then poured into one arm until the height of the alcohol column is 10 cm. The two liquids do not mix.

What is the difference in height between the top surface of the glycerin and the top surface of the alcohol? Suppose that the density of glycerin is 1260 kg/m3and the density of alcohol is 790 kg/m3.

Express your answer in two significant figures and include the appropriate units (in cm)

Answer:

ΔH ≅ 3.73 cm

Explanation:

The pressure inside a liquid is known as hydrostatic pressure and which is represent by the formula:

P = ρ × g × h

where;

ρ is the density of the fluid

g is the gravitational constant

h is the height from the surface

From the question above;

For glycerine; we have:

density of glycerine = 1260 kg/m³

gravitational constant = 9.8 m/s²

height = ???

∴

$P_{(g)= 1260kg/m^3}*9.8m/s^2*h_g$ ----- equation (1)

On the other hand for alcohol:

density of alcohol is given as = 790 kg/m³

gravitational constant = 9.8 m/s²

height = 10 cm

∴

$P_{(a)= 790kg/m^3*9.8m/s^2*10$ ----------- equation (2)

if we equate equation 1 and 2 together; we have

$P_{(g)= P_{(a)$

$1260kg/m^3}*9.8m/s^2*h_g = 790kg/m^3*9.8m/s^2*10cm$

Making $h_g$ the subject of the formula, we have :

$h_g= \frac{ 790kg/m^3*9.8m/s^2*10cm}{1260kg/m^3*9.8m/s^2}$

$h_g$ = 6.269 cm

The difference in the height denoted by ΔH can therefore be calculated as:

ΔH $= H_a-H_g$

ΔH $= 10cm - 6.269cm$

ΔH = 3.731 cm

ΔH ≅ 3.73 cm (to two significant figures)

5 0

2 years ago

Like many other organic solvents we will use this semester, ethanol is flammable and caution needs to be exercised when heating

laiz [17]

Answer:

See explanation

Explanation:

A flammable solvent refers to a solvent that catches fire easily. The precautions to be taken when working with flammable solvents are;

1) heat the solvent at a low to medium hot plate setting.

2) if you need to boil the solvent, use a condenser rather than a flask or beaker without a cover.

3) make sure that the hotplate is larger than the vessel containing the mixture that is being heated.

4) do not use strong oxidizing agents

5 0

2 years ago

Three of the reactions that occur when the paraffin of a candle (typical formula C21H44) burns are as follows:

Katarina [22]

Answer and Explanation:

The 3 reactions represented are

C₂₁H₄₄ + 32O₂ -----> 21CO₂ + 22H₂O

C₂₁H₄₄ + (43/2)O₂ -----> 21CO + 22H₂O

C₂₁H₄₄ + 11O₂ -----> 21C + 22H₂O

ΔH°(C₂₁H₄₄) = -476 KJ/mol, ΔH°(O₂) = 0KJ/mol, ΔH°(CO₂) = -393.5 KJ/mol, ΔH°(CO) = -99 KJ/mol, ΔH°(H₂O) = -292.74 KJ/mol, ΔH°(C) = 0KJ/mol

ΔH°f = ΔH°(products) - ΔH°(reactants)

For reaction 1,

ΔH°(products) = 21(ΔH°(CO₂)) + 22(ΔH°(H₂O)) = 21(-393.5) + 22(-292.74) = -14703.78 KJ/mol

ΔH°(reactants) = ΔH°(C₂₁H₄₄) = -476 KJ/mol

ΔH°f = -14703.78 - (-476) = - 14227.78 KJ/mol

For reaction 2,

ΔH°(products) = 21(ΔH°(CO)) + 22(ΔH°(H₂O)) = 21(-99) + 22(-292.74) = -8519.28 KJ/mol

ΔH°(reactants) = ΔH°(C₂₁H₄₄) = -476 KJ/mol

ΔH°f = -8519.28 - (-476) = - 8043.28 KJ/mol

For reaction 3,

ΔH°(products) = 21(ΔH°(C)) - 22(ΔH°(H₂O)) = 21(0) + 22(-292.74) = -6440.28 KJ/mol

ΔH°(reactants) = ΔH°(C₂₁H₄₄) = -476 KJ/mol

ΔH°f = -6440.28 - (-476) = - 5968.28 KJ/mol

b) To obtain the q for the combustion of 254g of paraffin, we convert the mass to moles.

Number of moles = mass/molar mass; molar mass of C₂₁H₄₄ = 296 g/mol

Number of moles = 254/296 = 0.858 mole

heat of reaction for the combustion of C₂₁H₄₄ when it is complete combustion, q = ΔH°(complete combustion, i.e. reaction 1) × number of moles = -14227.78 × 0.858 = -12207.435 KJ/mol

c) 8% of the mass of C₂₁H₄₄ undergoes incomplete combustion = 8% × 254 = 20.32g, in number of moles = 20.32/296 = 0.0686 mole

5% of the mass of C₂₁H₄₄ becomes soot = 5% × 254 = 12.7g, in number of moles = 12.7/296 = 0.0429 mole

The remaining paraffin undergoes complete combustion = 87% of 254 = 220.98g, in number of moles = 220.98 = 0.747 mole

q = sum of all the heat of reactions = (0.747 × -14227.78) + (0.0686 × -8043.28) + ( 0.0429 × -5968.28) = -11435.377 KJ

QED!!!

8 0

1 year ago

How many grams of NH3 are needed to provide the same number of molecules as in 0.35 g of SF6 ?

Strike441 [17]

Answer:

There are 0.041 g of NH₃ in the same number of molecules as in 0.35 g of SF₆.

Explanation:

Using the molar mass of the chemical formula SF₆ you can find the number of moles of molecules in 0.35 g of such substance. Then, using the molar mass of NH₃, you can find mass in grams corresponding to the same number of molecules.

1. Find the molar mass of SF₆:

Atom atomic mass number of atoms total mass in 1 mole

S 32.065 g/mol 1 32.065 g

F 18.998 g/mol 6 6 × 18.998 = 113.988 g

=====================

molar mass of SF₆ = 146.053 g/mol

2. Find the number of moles in 0.35 g of SF₆:

number of moles = mass in grams / molar mass

number of moles = 0.35 g / 146.053 g / mol = 0.0024 mol

3. Find the molar mass of NH₃:

Atom atomic mass number of atoms total mass 1 mole

N 14.007 g/mol 1 14.007 g

H 1.008 g/mol 3 3 × 1.008 g = 113.988 g

=====================

molar mass of NH₃ = 17.031 g/mol

4. Find the mass in 0.0024 mol of NH₃:

mass in grams = number of moles × molar mass

mass = 0.0024 mol × 17.031 g/mol ≈ 0.041 grams

5. Conclusion:

There are 0.041 g of NH₃ in the same number of molecules as in 0.35 g of SF₆.

6 0

1 year ago