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Llana [10]
3 years ago
12

Ca(OH)2 + H2SO4 ⟶2H2O + CaSO4 What volume of 1.45 M Ca(OH)2 is needed to react with 25.0 moles of H2SO4?

Chemistry
1 answer:
Shkiper50 [21]3 years ago
3 0

Answer:

We need 17.2 L of Ca(OH)2

Explanation:

Step 1: Data given

Concentration of Ca(OH)2 = 1.45 M

Moles of H2SO4 = 25.0 moles

Step 2: The balanced equation

Ca(OH)2 + H2SO4 ⟶2H2O + CaSO4

Step 3: Calculate moles Ca(OH)2

For 1 mol Ca(OH)2 we need 1 mol H2SO4 to produce 2 moles H2O and 1 mol CaSO4

For 25.0 moles H2SO4 we'll need 25.0 moles Ca(OH)2 to produce 50 moles H2O and 25.0 moles CaSO4

Step 4: Calculate volume of Ca(OH)2

Volume Ca(OH)2 = moles Ca(OH)2 / concentration Ca(OH)2

Volume Ca(OH)2 = 25.0 moles / 1.45 M

Volume Ca(OH)2 = 17.2 L

We need 17.2 L of Ca(OH)2

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Type the correct answer in the box.
IgorLugansk [536]

Answer:

a= <em>In scientific notation</em>

6.96000×10⁵ Km

b =<em>In  expanded notation</em>

0.00019 mm

Explanation:

Given data:

Radius of sun = 696000 Km

size of bacterial cell = 1.9 ×10⁻⁴ mm

Radius of sun in scientific notation = ?

Size of bacterial cell in expanded notation = ?

Solution:

Scientific notation is the way to express the large value in short form.

The number in scientific notation have two parts.

. The digits (decimal point will place after first digit)

× 10 ( the power which put the decimal point where it should be)

for example the number 6324.4 in scientific notation will be written as = 6.3244 × 10³

Radius of sun:

696000 Km

<em>In scientific notation</em>

6.96000 × 10⁵ Km

The expanded notation is standard notation of writing the numerical values which is normal way. The numbers are written as they are, without the power of 10.

Size of bacterial cell:

1.9 ×10⁻⁴ mm

<em>In  expanded notation</em>

1.9/ 10000 = 0.00019 mm

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3 years ago
Evaluate the Lewis structures below and pick the statements that best describes the accuracy of each of the structures. There ma
Artyom0805 [142]

Answer:

The structures shown by dots and lines to give the exact number of electrons in the outer most shell is explained by Lewis Structures.

Explanation:

Lewis structures are those structures in which the diagram is shown using the electron representation. They are easy to understand as the diagram completely depicts where the electrons are shared and where they are transferred. The diagram also explains where there is a single bond and where there is a di covalent bond or tri covalent bond explaining where the single , double or triple electron pair is shared. The electrons are shown by dots or lines.

For example CCl₄ can be shown as follows

                  ..

               .. Cl..

  ..                               ..

..Cl..----------C----------..Cl..

                   ..

                .. Cl..

The picture shows that each chlorine has six electrons in its outer shell and then a pair of electron is shared with carbon forming a single covalent bond.

Similarly methane CH4 can also be shown.

The hydrogen has one electron and it shares an electron from carbon stabilising itself forming methane.

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3 years ago
In Chemistry, to be classified as an organic substance, a substance must contain
Natasha_Volkova [10]
I'm going on a limb here, but Carbon is a definite. <span />
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3 years ago
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Alex73 [517]
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4 0
3 years ago
Assume that the density and heat of combustion of E85 can be obtained by using 85 % of the values for ethanol and 15 % of the va
Alika [10]

Answer:

87.4 J

Explanation:

The density of the gasoline is 0.70 g/mL, and the density of the ethanol is 0.79 g/mL. The heat combustions (the heat released in a combustion reaction) are 5,400 kJ/mol for gasoline, and 1,370 kJ/mol for ethanol.

For 3.5 L of E85, the volumes of gasoline and ethanol are:

Vgasoline = 0.15 * 3.5 = 0.525 L = 5.25x10⁻⁴ mL

Vethanol= 0.85 * 3.5 = 2.975 L = 2.975x10⁻³ mL

The mass of gasoline and ethanol presented in that sample of E85 is the volume multiplied by the density:

mgasoline = 5.25x10⁻⁴ * 0.70 = 3.675x10⁻⁴ g

methanol = 2.975x10⁻³ * 0.79 = 2.35025x10⁻³ g

The number of moles for each substance is it mass divided by its molar mass. The molar masses are 114 g/mol for gasoline, and 46 g/mol for ethanol:

ngasoline = 3.675x10⁻⁴/114 = 3.224x10⁻⁶ mol

nethanol = 2.35025x10⁻³ /46 = 5.109x10⁻⁵ mol

The energy released is the heat combustion multiplied by the number of moles, so:

Egasoline = 5,400 * 3.224x10⁻⁶ = 0.0174 kJ = 17.4 J

Eethanol = 1,370 * 5.109x10⁻⁵ = 0.07 kJ = 70 J

So, the energy released by the E85 is the sum of the energy released by ethanol and gasoline:

The energy released by E85 = 87.4 J

8 0
3 years ago
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