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3 years ago

The standard enthalpy of formation of Ca2+ (g) (∆H = 1934 kJ/mol)

Chemistry

1 answer:

Alex73 [517]3 years ago

6 0

Answer:

see explanation

Explanation:

The process of ionization to produce cations is endothermic. For formation of Ca⁺² two ionization steps need be illustrated as follows...

1st ionization step: Ca° + 590Kj => Ca⁺ + e⁻

2nd ionization step: Ca⁺ + 1151Kj => Ca⁺² + e⁻

__________________________________-

Net Ionization Rxn: Ca° + 1741Kj => Ca⁺² + 2e⁻

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How many Ne atoms are contained in 32.0 g of the element?

Svetlanka [38]

Mass atomic of Ne=20.18 u
Therefore:
molar mass=20.18 g/1 mol

1 mole=6.022*10²³ particles (atoms or molecules)

Then: 6.022*10²³ atoms are contained in 20.18g

Now, We can solve this problem by the three rule.

6.022*10²³ atoms-------------------20.18 g
x------------------------------------------32 g

x=(6.022*10²³ atoms * 32 g)/20.18 g=9.55*10²³ atoms.

Answer: 9.55*10²³ Ne atoms are contained in 32 g of the element.

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3 years ago

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Goshia [24]

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2 years ago

Valeric acid, HC5H9O2 (Ka = 1.5 ✕ 10−5), is used in the manufacture of magnesium valerate, a nerve-calming agent. What is the hy

Studentka2010 [4]

Answer:

$[H^+]=0.00332M$

Explanation:

Hello,

In this case, considering the dissociation of valeric acid as:

$HC_5H_9O_2 \rightleftharpoons C_5H_9O_2 ^-+H^+$

Its corresponding law of mass action is:

$Ka=\frac{[H^+][C_5H_9O_2^-]}{[HC_5H_9O_2]}$

Now, by means of the change $x$ due to dissociation, it becomes:

$Ka=\frac{(x)(x)}{0.737-x}=1.5x10^{-5}$

Solving for $x$ we obtain:

$x=0.00332M$

Thus, since the concentration of hydronium equals $x$ , the answer is:

$[H^+]=x=0.00332M$

Best regards.

7 0

3 years ago

When you need to produce a variety of diluted solutions of a solute, you can dilute a series of stock solutions. A stock solutio

8090 [49]

Answer:

Volume of stock solution needed = 6.0299 mL

Explanation:

<u> </u>Dilution consists of lowering the amount of solute per unit volume of solution. It is achieved by adding more diluent to the same amount of solute.

This is deduced when thinking that both the dissolution at the beginning and at the end will have the same amount of moles.

M1 = 6.01 M stock solution concentration

M2 = 0.3624 M diluted solution concentration

V2 =100 mL diluted solution volume

V1 = ? stock solution volume

M1 * V1 = M2 * V2

$V1=\frac{M2*V2}{M1} =\frac{0.3624M*100mL}{6.01M} =6.0299 mL$

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zhenek [66]

CLOUDS would be one of the interacting parts in a weather system.

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