What should each experiment only have one of?

How many milliliters of 0.50 M KOH are needed

spayn [35]

Answer:

Option D. 30 mL.

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

HNO3 + KOH —> KNO3 + H2O

From the balanced equation above,

The mole ratio of the acid, nA = 1

The mole ratio of the base, nB = 1

Step 2:

Data obtained from the question. This include the following:

Volume of base, KOH (Vb) =.?

Molarity of base, KOH (Mb) = 0.5M

Volume of acid, HNO3 (Va) = 10mL

Molarity of acid, HNO3 (Ma) = 1.5M

Step 3:

Determination of the volume of the base, KOH needed for the reaction. This can be obtained as follow:

MaVa / MbVb = nA/nB

1.5 x 10 / 0.5 x Vb = 1

Cross multiply

0.5 x Vb = 1.5 x 10

Divide both side by 0.5

Vb = (1.5 x 10) /0.5

Vb = 30mL

Therefore, the volume of the base, KOH needed for the reaction is 30mL.

3 0

2 years ago

Write the limiting forms (or Canonical forms) of the following ions:

nekit [7.7K]

Answer:

Canonical structures of a chemical specie explain its observed properties from a valence bond theory perspective.

Explanation:

Resonance is a valence bond concept introduced by Linus Pauling to explain the observed properties of certain chemical species such as bond lengths, bond angles, bond order , etc.

There are certain chemical species for which a single chemical structure does not suffice in explaining its observed properties. For instance, the bond order in CO3^2- is about 1.33. Its bond length, shows that the C-O bond present in CO3^2- is neither a pure C-O single bond nor a pure C-O double bond. Hence the structure of CO3^2- is 'somewhere in between' three contributing canonical structures as shown in the image attached to this answer. The resonance structures of NO3^- are also shown.

3 0

2 years ago

The rate of effusion of an unknown gas was measured and found to be 11.9 mL/min. Under identical conditions, the rate of effusio

iren2701 [21]

Answer : The correct option is, (B) $CO_2$

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

$R\propto \sqrt{\frac{1}{M}}$

or,

$(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}$ ..........(1)

where,

$R_1$ = rate of effusion of unknown gas = $11.9\text{ mL }min^{-1}$

$R_2$ = rate of effusion of oxygen gas = $14.0\text{ mL }min^{-1}$

$M_1$ = molar mass of unknown gas = ?

$M_2$ = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the above formula 1, we get:

$(\frac{11.9\text{ mL }min^{-1}}{14.0\text{ mL }min^{-1}})=\sqrt{\frac{32g/mole}{M_1}}$

$M_1=44.2g/mole$

The unknown gas could be carbon dioxide $(CO_2)$ that has approximately 44 g/mole of molar mass.

Thus, the unknown gas could be carbon dioxide $(CO_2)$

5 0

3 years ago

How important is the resonance structure shown here to the overall structure of carbon dioxide?

Tcecarenko [31]

1. Resonance structures are a better description of a Lewis dot structure .

2. Best resonance structure is the one with the least formal charge.

<u>EXPLAINATION OF RASONANCE STRUCTURE OF CARBON DIOXIDE</u>

1.Carbon dioxide has three resonance structures .

2. The CO2 molecule has a total of 16 valence electrons ,

1C = 4 electrons

2O= 12 electrons

<u>three resonance structures for CO2</u>

1. The atoms in all three resonance structures have full octets;

structure 1 will be more stable because it has no separation of charge.

2. Structures 2 and 3 show charge separation caused by the presence of formal charges on both oxygen atoms.

6 0

2 years ago

Suppose a system of two particles, represented by circles, have the possibility of occupying energy states with 0, 10, or 20 J.

MrMuchimi

Answer:

Explanation:

The possible energy states for the particles are 0, 10 and 20 J.

The constraint in the system is that the total energy of the particles must be 20 J.

One given configuration where the total energy is 20 J is if both the particles occupy the 10 J state.

Hence, (10;10) is the given configuration.

Another possibility is if one of the particle is in 0 J state and another is in 20 J state. Hence, the system has a total energy of 0+20 = 20 J.

Hence, the possible configuration can be written as (0;20) or (20;0) which are energetically equivalent to the given configuration. Note that if the circles are indistinguishable, then the configuration (0,20) and (20,0) is the same thing.

4 0

2 years ago