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3 years ago

What does the suffix ite stand for

Chemistry

1 answer:

insens350 [35]3 years ago

8 0

Answer: The -ite suffix is used on the oxyanion with one oxygen atom fewer (like sulfite SO32- or nitrite NO2-).

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Gold has a high potential and potassium has a low chemical potential.

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Gold potential is the correct answer

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2 years ago

A naturally occurring, inorganic solid with an orderly crystalline structure and a definite chemical composition is _____.

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Naturally occurring, inorganic, crystalline structure is a mineral

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Read 2 more answers

Which of the following reactions is balanced?

77julia77 [94]

Answer:

B. CaCl₂ + H₂CO₃ → CaCO₃ + 2HC

Explanation:

A balanced reaction has the same number of atoms in the both sides of the reaction. In the options:

A. CaCl₂ + H₂CO₃ → 2CaCO₃ + HCI

In this reaction there is 1 Ca in reactants and 2 in products -The reaction is unbalanced-

There is 1 Ca is both sides, 2Cl, 2H, 1C and 3 Oxygens -The reaction is balanced

C. CaCl₂ + 2H₂CO₃ → CaCO₃ + HCI

There is 1 Ca in both sides but 2Cl in reactants and 1 in Cl -The reaction is unbalanced-

D. 2CaCl₂ + H₂CO₃ →CaCO₃ + HCI

There are 2 Ca in reactants and 1 in Ca -The reaction is unbalanced-

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3 years ago

Which layer of the atmosphere is between the mesosphere and the exosphere?

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C. Thermosphere is the correct answer

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2 years ago

Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolv

Triss [41]

Answer:

1. $3.57\times 10^{-4}$ was the $K_a$ value calculated by the student.

2. $5.93\times 10^{-4}$ was the $K_b$ of ethylamine value calculated by the student.

Explanation:

The $pH$ value of Aspirin solution = 2.62

$pH=-\log[H^+]$

$[H^+]=10^{-2.62}=0.00240 M$

Moles of s asprin = $\frac{2.00 g}{180 g/mol}=0.01111 mol$

Volume of the solution = 0.600 L

The initial concentration of Aspirin = c = $\frac{0.01111 mol}{0.600 L}=0.0185 M$

$HAs\rightleftharpoons As^-+H^+$

initially

c 0 0

At equilibrium

(c-x) x x

The expression of dissociation constant :

$K_a=\frac{[As^-][H^+]}{[HAs]}$ :

$K_a=\frac{x\times x }{(c-x)}$

$=\frac{0.00240 M\times 0.00240 M}{(0.0185-0.00240 )}$

$K_a=3.57\times 10^{-4}$

$3.57\times 10^{-4}$ was the $K_a$ value calculated by the student.

The $pH$ value of ethylamine = 11.87

$pH+pOH=14$

$pOH=14-11.87=2.13$

$pOH=-\log[OH^-]$

$[OH^-]=10^{-2.13}=0.00741 M$

The initial concentration of ethylamine = c = 0.100 M

$C_2H_5NH_2+H_2O\rightleftharpoons C_2H_5NH_3^{+}+OH^-$

initially

c 0 0

At equilibrium

(c-x) x x

The expression of dissociation constant :

$K_b=\frac{[C_2H_5NH_3^{+}][OH^-]}{[C_2H_5NH_2]}$ :

$K_b=\frac{x\times x}{(c-x)}$

$=\frac{0.00741\times 0.00741}{(0.100-0.00741)}$

$K_b=5.93\times 10^{-4}$

$5.93\times 10^{-4}$ was the $K_b$ of ethylamine value calculated by the student.

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3 years ago