Which of these structures are found in both plant and animal cells?

1) Aluminum sulphate can be made by the following reaction: 2AlCl3(aq) + 3H2SO4(aq) Al2(SO4)3(aq) + 6 HCl(aq) It is quite solubl

kolezko [41]

Answer:

88.9%

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2AlCl3(aq) + 3H2SO4(aq) —> Al2(SO4)3(aq) + 6HCl(aq)

Step 2:

Determination of the masses of AlCl3 and H2SO4 that reacted and the mass of Al2(SO4)3 produced from the balanced equation.

Molar mass of AlCl3 = 27 + (35.5x3) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Molar mass of H2SO4 = (2x1) + 32 + (16x4) = 98g/mol

Mass of H2SO4 from the balanced equation = 3 x 98 = 294g

Molar mass of Al2(SO4)3 = (27x2) + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96] = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4 to produce 342g of Al2(SO4)3.

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4.

Therefore, 25g of AlCl3 will react with = (25 x 294)/267 = 27.53g of H2SO4.

From the calculations made above, we see that only 27.53g out 30g of H2SO4 given were needed to react completely with 25g of AlCl3.

Therefore, AlCl3 is the limiting reactant and H2SO4 is the excess.

Step 4:

Determination of the theoretical yield of Al2(SO4)3.

In this case we shall be using the limiting reactant because it will produce the maximum yield of Al2(SO4)3 since all of it is used up in the reaction.

The limiting reactant is AlCl3 and the theoretical yield of Al2(SO4)3 can be obtained as follow:

From the balanced equation above,

267g of AlCl3 reacted to produce 342g of Al2(SO4)3.

Therefore, 25g of AlCl3 will react to produce = (25 x 342) /267 = 32.02g of Al2(SO4)3.

Therefore, the theoretical yield of Al2(SO4)3 is 32.02g

Step 5:

Determination of the percentage yield of Al2(SO4)3.

This can be obtained as follow:

Actual yield of Al2(SO4)3 = 28.46g

Theoretical yield of Al2(SO4)3 = 32.02g

Percentage yield of Al2(SO4)3 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 28.46/32.02 x 100

Percentage yield = 88.9%

Therefore, the percentage yield of Al2(SO4)3 is 88.9%

3 0

4 years ago

Antimony has 2 stable isotopes 121-sb mass 120.9038 and 123-sb mass 122.9042 amu. calculate the percent abundances of these isot

iragen [17]

The atomic mass of an element is a result of the weighted average of the masses of its corresponding isotopes.

We are given: 
Sb-121: mass of 120.9038 amu, x abundance 
Sb-123: mass of 122.9042 amu, 1-x abundance 

To be able to calculate the atomic mass of antimony, we multiply the percent abundances of the isotopes by their respective atomic masses. Then, we add,

Atomic mass = (Atomic Mass of Sb-121) (% Abundance) + (Atomic Mass of Sb-123) (% Abundance )

121.760 amu = (120.9038 amu)(x) + (122.9042 amu)(1 -x)

Solving for x,
120.9038x + 122.9042 -122.9042x = 121.760
x = 0.57096 or 57.096% 
1-x = 1-0.57096 = 0.42904 or 42.90%

6 0

4 years ago

A gas has a pressure of 500.0 Torr at 35.0 °C. What is the temperature at standard pressure?

Aleks04 [339]

Answer:

466.65 K

Explanation:

Using the formula of Pressure law,

P/T = P'/T'.................... Equation 1

Where P = Intial pressure of gas, T = Initial temperature of gas, P' = Final pressure of gas, T' = Final temperature of gas.

make T' The subeject of equation 1

T' = (P'T)/P................... Equation 2

From the question,

Given: P = 500 Torr = (500×133.322) N/m² = 66661 N/m², T = 35°C = (35+273) = 308 K, P' = 101000 N/m²( Standard pressure)

Substitute these values into equation 2

T' = (101000×308)/66661

T' = 466.65 K

5 0

3 years ago

Explain how the valence electrons are related to the element's period.

AlekseyPX

Answer:

The number of valence electrons in an atom is reflected by its position in the periodic table of the elements (see the periodic table in the Figure below). Across each row, or period, of the periodic table, the number of valence electrons in groups 1–2 and 13–18 increases by one from one element to the next

5 0

3 years ago

The ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. in t

Debora [2.8K]

The complete balanced chemical equation is:
4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g)

In statement form: 4mol NH3 reacts with 5 mol O2 to produce 6 mol H2O

First let us find for the limiting reactant:
>molar mass NH3 = 17 g/mol
moles NH3 = 54/17 = 3.18 mol NH3
This will react with 3.18*5/4 = 3.97 mol O2

>molar mass O2 = 32g/mol
moles O2 = 54/32 = 1.69 mol O2
We have insufficient O2 therefore this is the limiting reactant

From the balanced equation:
For every 5.0 mol O2, we get 6.0 mol H2O, therefore

moles H2O formed = 1.69 mol O2 * 6/5 = 2.025 mol
Molar mass H2O = 18g/mol
mass H2O formed = 2.025*18 = 36.45 grams H2O produced

8 0

4 years ago

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