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irina1246 [14]
3 years ago
11

According to the arrhenius concept, if HNO3, were dissolved in water, it would act as?: a. Proton acceptor b. A base c. A source

of hydroxide ions d. An acid
Chemistry
2 answers:
Alexxandr [17]3 years ago
5 0
Arrhenius concept just states that if a solution dissociates and forms H+ ions then its and acid, but if it dissociates and forms OH- than it's a base.

So the answer to your question would be D. An Acid
Alenkasestr [34]3 years ago
3 0

Answer: d. An acid

Explanation:

According to Arrhenius concept, a base is defined as a substance which donates hydroxide ions (OH^-) when dissolved in water and an acid is defined as a substance which donates hydrogen ions (H^+) in water.

A substance that increases the hydrogen ion concentration of a solution is an Arrhenius acid.

HNO_3\rightarrow NO_3^-+H^+

Thus as HNO_3 donates a proton , it is considered as an Arrhenius acid.

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What would be the voltage (Ecell) of a voltaic cell comprised of Cd(s)/Cd2+(aq) and Zr(s)/Zr4+(aq) if the concentrations of the
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Answer:

1.05 V

Explanation:

Since;

E°cell= E°cathode- E°anode

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Equation of the process;

2Zr(s) + 4Cd^2+(aq) ---->2Zr^4+(aq) + 4Cd(s)

n= 8 electrons transferred

From Nernst's equation;

Ecell = E°cell - 0.0592/n log Q

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Ecell= E°cell= 1.05 V

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3 years ago
The reducing agent in the reaction described in Fe + 2HCl → FeCl2 + H2 is A. H2. B. Fe. C. HCl. D. FeCl2.
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The reducing agent is b). Fe
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3 years ago
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Using the periodic table,<br> choose the more reactive nonmetal. Br or as
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Answer:

Br

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7 0
3 years ago
A space air is at a temperature of 75 oF, and the relative humidity (RH) is 45%. Using calculations, find: (a) the partial press
earnstyle [38]

Answer:

A) Partial Pressure of dry air = 13.32 KPa

Partial Pressure of water vapour = 1.332 KPa

B) Humidity ratio; X = 0.0691

C) V_p = 0.8384 m³/Kg

Explanation:

A) We are given;

Temperature = 75°F

Relative Humidity = 45%

Now,to calculate the partial pressure, we will use the relationship;

Relative Humidity = (Partial Pressure/Vapour Pressure) × 100%

Making partial pressure the subject;

Partial Pressure = Relative Humidity × Vapour Pressure/100%

From the first table attached, at temperature of 75°F, the vapor pressure is 29.6 × 10^(-3) bar = 29.6 KPa

Thus;

Partial Pressure of dry air = (45 × 29.6)/100

Partial Pressure of dry air = 13.32 KPa

From online values, vapour pressure of water vapour at 75°F = 2.96 KPa

Thus;

Partial Pressure of water vapour = (45 × 2.96)/100 = 1.332 KPa

B) humidity ratio of moist air is given as;

X = 0.62198 pw / (pa - pw)  

where;

pw = partial pressure of the water vapor in moist air

pa = atmospheric pressure of the moist air

Thus;

X = (0.62198 × 1.332)/(13.32 - 1.332)  

X = 0.0691

C) Formula for moist air specific volume is;

V_p = (1 + (xRw/Ra) × RaT/p

Where;

V_p is specific volume

T is temperature = 75°F = 297.039 K

p is barometric pressure which in this case is standard sea level pressure = 101.325 KPa

pw is partial pressure of the water vapor in moist air = 1.332 KPa

Rw is individual gas constant for water = 0.4614 KJ/Kg.K

Ra is individual gas constant for air = 0.2869 KJ/Kg.K

V_p = (1 + (0.0691 * 0.4614/0.2869)) × 0.286.9 * 297.039/101.325

V_p = 0.8384 m³/Kg

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