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2 years ago

3. Write DISSOCIATION or IONIZATION equations for the following chemicals:

Chemistry

1 answer:

Romashka [77]2 years ago

8 0

Answer:

a) HNO3 -> H+ + NO3- disassociation of Nitric Acid; to yield a Nitrate ion and a Proton, H+, or as a Hydronium ion H3O+

b) H2S04 -> Disassociation of Sulfuric Acid; simple way- 2H+ + SO4- -

c) H2S hydrogen sulphide in water is an acid; thus H+ HS- disassociation.

d) NaOH -> dissociation of Na+ + OH-; this is complete; sodium hydroxide is deliquescent, meaning it will draw water - EVEN from the air! Strong Base

e) Na2CO3 -> 2Na+ CO3- - Ionization of sodium carbonate - a salt

f) Na2S04 -> 2Na+ + SO4 - - ionization of sodium sulphate - a salt

g) NaCl -> Na+ + Cl- ionization of the salt, Sodium Chloride

Explanation:

Salts ionize at different rates; acids or bases dissociate; these are mostly strong acids and NaOH, a strong base.

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If an atom has 15 protons, 11 neutrons, and 16 electrons, what is the atom's electrical charge? A. -5 B. -1 C. +1 D. +5

Helen [10]

Answer:

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Explanation:

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2 years ago

There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In

KIM [24]

Answer:

ΔH = -470.4kJ

Explanation:

It is possible to sum 2 or more reactions to obtain the ΔH of the reaction you want to study (Hess's law). Using the reactions:

1. CaC2(s) + 2H2O(l) → C2H2(g) + Ca(OH)2(s)ΔH = −414kJ

2. 6C2H2(g) + 3CO2(g) + 4H2O(g) → 5CH2CHCO2H(g)ΔH = 132kJ

6 times the reaction 1.

6CaC2(s) + 12H2O(l) → 6C2H2(g) + 6Ca(OH)2(s)ΔH = −414kJ*6 = -2484kJ

This reaction + 2:

6CaC2(s) + 3CO2(g) + 16H2O(l) → + 6Ca(OH)2(s) + 5CH2CHCO2H(g) ΔH = -2484kJ + 132kJ = -2352kJ

As we want to calculate the net change enthalpy in the formation of just 1 mole of acrylic acid we need to divide this last reaction in 5:

6/5CaC2(s) + 3/5CO2(g) + 16/5H2O(l) → + 6/5Ca(OH)2(s) + CH2CHCO2H(g) ΔH = -2352kJ / 5

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2 years ago

Match the following numbers with the correct number of significant digits they possess. Answers may repeat. (4 points)

mezya [45]

(B. 3) 172 All nonzero digits are significant.

(A. 4) 450.0 x 10^3 Trailing zeroes after the decimal point are significant.

(A. 4) 3427 All nonzero digits are significant.

(B. 3) 0.0000455 Leading zeroes are not significant.

(B. 3) 0.00456 Leading zeroes are not significant.

(C. 5) 2205.2 Zeroes between nonzero digits are significant.

(C. 5) 107.20 Trailing zeroes after the decimal point are significant.

(B. 3) 0.0473 Leading zeroes are not significant.

8 0

3 years ago

Which element has exactly ﬁve electrons in the highest principal energy level (the outer shell)?

statuscvo [17]

This question is asking for an element with 5 valence electrons. Just go to the row it is in (excluding transition metals) and count over.
The answer would be c. P

7 0

3 years ago

Read 2 more answers

For the reaction, 2A + B → 3C, rate of disappearance of A is 0.3 M/s in a certain time interval.What is the average rate of appe

andrezito [222]

Answer:

E) rate of appearance of C = 0.45 M/s

rate of the reaction = 0.15 M/s

Explanation:

2A + B → 3C

Writing rate law for the reaction:

<u>Rate of reaction</u> = - $\frac{1}{2}$ $\frac{d[A]}{dt}$ = - $\frac{d[B]}{dt}$ = $\frac{1}{3}$ $\frac{d[C]}{dt}$ → equation 1

Given that the rate of disappearance of A is 0.3 M/s

⇒ - $\frac{d[A]}{dt}$ = 0.3 M/s

⇒Rate of reaction = - $\frac{1}{2}$ $\frac{d[A]}{dt}$ = $\frac{1}{2}$ ×0.3 M/s

⇒<u>Rate of reaction = 0.15 M/s</u>

From equation 1, $\frac{d[C]}{dt}$ = - $\frac{3}{2}$ $\frac{d[A]}{dt}$ = $\frac{3}{2}$ ×0.3 M/s

⇒ $\frac{d[C]}{dt}$ = 0.45 M/s

or <u>the rate of appearance of C = 0.45 M/s</u>

7 0

2 years ago