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mel-nik [20]
2 years ago
9

How many ml of 0.5M of HNO3 would be needed to react with 85ml of 0.75M of KOH

Chemistry
1 answer:
olganol [36]2 years ago
6 0

Answer: 127.5ml

Explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HNO_3

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is KOH.

We are given:

n_1=1\\M_1=0.5M\\V_1=?mL\\n_2=1\\M_2=0.75M\\V_2=85mL

Putting values in above equation, we get:

1\times 0.5\times V_1=1\times 0.75\times 85\\\\V_1=127.5mL

Thus 127.5 ml of 0.5M of HNO3 would be needed to react with 85ml of 0.75M of KOH

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At which temperature do the molecules of an ideal gas have 3 times the kinetic energy they have at 32of?
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Answer:

  • 820 K

Explanation:

As per Boltzman equation, <em>kinetic energy (KE)</em> is in direct relation to the <em>temperature</em>, measured in absolute scale Kelvin.

  • KE α T.

Then, <em>the temperature at which the molecules of an ideal gas have 3 times the kinetic energy they have at any given temperature will be </em><em>3 times</em><em> such temperature.</em>

So, you must just convert the given temperature, 32°F, to kelvin scale.

You can do that in two stages.

  • First, convert 32°F to °C. Since, 32°F is the freezing temperature of water, you may remember that is 0°C. You can also use the conversion formula: T (°C) = [T (°F) - 32] / 1.80

  • Second, convert 0°C to kelvin:

         T (K) = T(°C) + 273.15 K= 273.15 K

Then, <u>3 times</u> gives you: 3 × 273.15 K = 819.45 K

Since, 32°F has two significant figures, you must report your answer with the same number of significan figures. That is 820 K.

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What is the limiting reactant when 4 mol P4 and 4 mol S8 react.
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Taking into account the reaction stoichiometry, P₄ will be the limiting reagent.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

8 P₄ + 3 S₈ → 8 P₄S₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

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<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 3 moles of S₈ reacts with 8 moles of P₄, 4 moles of S₈ reacts with how many moles of P₄?

moles of P_{4} =\frac{4 moles of S_{8} x8 moles of P_{4} }{3 moles of S_{8}}

<u><em>moles of P₄= 10.667 moles</em></u>

But 10.667 moles of P₄ are not available, 4 moles are available. Since you have less amount of moles than you need to react with 4 moles of S₈, P₄ will be the limiting reagent.

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The arrangement of the solutions based on their absorption from highest frequency to lowest frequency :

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