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3 years ago

An unknown solution has a pH of 2 . How would you classify this solution?

Chemistry

2 answers:

o-na [289]3 years ago

7 0

Answer is: B. acidic.

pH (potential of hydrogen) is a numeric scale used to specify the acidity or basicity of an aqueous solution:

When pH is less than seven (pH<7), solution is acidic.

When is equal seven (pH = 7), solution is neutral.

When pH is greater than seven (pH > 7), solution is basic.

For example, hydrochloric acid( HCl) is acidic, it ionizes in water solution and gives the hydrogen cations and chloride anions (Cl⁻):

HCl(aq) → Cl⁻(aq) + H⁺(aq).

marin [14]3 years ago

7 0

B. ACIDIC :) pH<7 is considered to be acidic

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In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for

inessss [21]

Answer:

The standard cell potential of the reaction is 0.78 Volts.

Explanation:

$Cu^{2+}(aq)+Fe(s)\rightarrow Cu(s)+Fe^{2+}(aq)$

Reduction at cathode :

$Cu^+(aq)+2e^-\rightarrow Cu(s)$

Reduction potential of $Cu^{2+}$ to Cu= $E^o_{1}=0.34 V$

Oxidation at anode:

$Fe(s)\rightarrow Fe^{2+}(aq)+2e^-$

Reduction potential of $Fe^{2+}$ to Fe= $E^o_{2}=-0.44 V$

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{red,cathode}-E^o_{red,anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.34V -(-0.44 V)=0.78 V$

The standard cell potential of the reaction is 0.78 Volts.

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hichkok12 [17]

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2 years ago

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nikklg [1K]

Answer:

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Explanation:

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2 years ago

Read 2 more answers

One mole of an ideal gas is expanded from a volume of 1.00 liter to a volume of 8.00 liters against a constant external pressure

Leya [2.2K]

Answer : The work done on the surroundings is, 709.1 Joules.

Explanation :

The formula used for isothermally irreversible expansion is :

$w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)$

where,

w = work done

$p_{ext}$ = external pressure = 1.00 atm

$V_1$ = initial volume of gas = 1.00 L

$V_2$ = final volume of gas = 8.00 L

Now put all the given values in the above formula, we get :

$w=-p_{ext}(V_2-V_1)$

$w=-(1.00atm)\times (8.00-1.00)L$

$w=-7L.atm=-7\times 101.3J=-709.1J$

The work done by the system on the surroundings are, 709.1 Joules. In this, the negative sign indicates the work is done by the system on the surroundings.

Therefore, the work done on the surroundings is, 709.1 Joules.

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3 years ago