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3 years ago

What volume of 0.500 M HNO3(aq) must completely react to neutralize 100.0 milliliters of 0.100 M KOH(aq)?

1 answer:

6 0

KOH+ HNO3--> KNO3+ H2O
From this balanced equation, we know that 1 mol HNO3= 1 mol KOH (keep in mind this because it will be used later).

We also know that 0.100 M KOH aqueous solution (soln)= 0.100 mol KOH/ 1 L of KOH soln (this one is based on the definition of molarity).

First, we should find the mole of KOH:
100.0 mL KOH soln* (1 L KOH soln/ 1,000 mL KOH soln)* (0.100 mol KOH/ 1L KOH soln)= 1.00*10^(-2) mol KOH.

Now, let's find the volume of HNO3 soln:
1.00*10^(-2) mol KOH* (1 mol HNO3/ 1 mol KOH)* (1 L HNO3 soln/ 0.500 mol HNO3)* (1,000 mL HNO3 soln/ 1 L HNO3 soln)= 20.0 mL HNO3 soln.

The final answer is (2) 20.0 mL.

Also, this problem can also be done by using dimensional analysis.

Hope this would help~

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Graph D.

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This is

V = K T, where K is the constant of proportionallity.

You can also write it in the form

V1 / V2 = T1/ Ts

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3 years ago

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Explanation:

In a magnetic field, the radius of the charged particle is as follows.

r = $\frac{mv}{qB}$

where, m = mass, v = velocity

q = charge, B = magnetic field

Therefore, q will be calculated as follows.

q = $\frac{mv}{rB}$

= $\frac{6.6 \times 10^{-27} \times 5.3 \times 10^{5}}{0.014 m \times 0.78 T}$

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= $3.2 \times 10^{-19} \times \frac{1.0 e}{1.6 \times 10^{-19}C}$

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Thus, we can conclude that the charge of the ionized atom is +2e.

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2 years ago

12. Describe the results of a chemical change. List four indicators of chemical change.

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1. Color Change

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2 years ago

Positive Deviation from Raoult's Law occurs when the vapour pressure of component is greater than what is expected in Raoult's L

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Answer:

All of the above.

Explanation:

In positive deviation from Raoult's Law occur when the vapour pressure of components is greater than what is expected value in Raoult's law.

When a solution is non ideal then it shows positive or negative deviation.

Let two solutions A and B to form non- ideal solutions.let the vapour pressure of component A is $P_A$ and vapour pressure of component B is $P_B$ .

$P^0_A$ = Vapour pressure of component A in pure form

$P^0_B$ = Vapour pressure of component B in pure form

$x_A$ =Mole fraction of component A

$x_B=$ =Mole fraction of component B

The interaction between A- B is less than the interaction A- A and B-B interaction.Therefore, the escaping tendency of liquid molecules in mixture is greater than the escaping tendency in pure form.Hence, the vapour pressure of a mixture is greater than the initial value of vapour pressure.

$P_A >P^0_A\cdot x_A$ , $P_B>P^0_B\cdot x_B$

Therefore, $P_A+P_B >P^0_A\cdot x_A+P^0_B \cdot x_B$

Therefore, the enthalpy of mixing is greater than zero and change in volume is greater than zero.

Hence, option a,b,c and d are true.

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3 years ago