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Arisa [49]
3 years ago
15

• 1) The pressure of a sample of gas in a 2.00-L container is 876 mmHg.

Chemistry
1 answer:
kifflom [539]3 years ago
5 0

Answer:

4380 mmHg

Explanation:

Boyle's Law can be used to explain the relationship between pressure and volume of an ideal gas. The pressure is inversely related to volume, so if volume decrease the pressure will increase. It can be expressed in the equation as:

P1V1=P2V2

In this question, the first condition is 2L volume and 876 mmHg pressure. Then the system changed into the second condition where the volume is 400ml and the pressure is unknown. The pressure will be:

P1V1= P2V2

876 mmHg * 2L = P2 * 400ml /(1000ml/L)

P2= 876 mmHg * 2L / 0.4L

P2= 4380 mmHg

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The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way:
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Explanation:

(a)  As the given chemical reaction equation is as follows.

           OCl^{-} + I^{-} \rightarrow OI^{-1} + Cl^{-1}

So, when we double the amount of hypochlorite or iodine then the rate of the reaction will also get double. And, this reaction is "first order" with respect to hypochlorite and iodine.

Hence, equation for rate law of reaction will be as follows.

              Rate = K \times [OCl^{-}] \times [l^{-}]

(b)  Since, the rate equation is as follows.

                    Rate = K [OCl^{-}][l^{-}]

Let us assume that ([OCl^{-}] = [l^{-}])

Putting the given values into the above equation as follows.

             1.36 \times 10^{-4} = K \times (1.5 \times 10^{-3})^2

            1.36 \times 10^{-4} = K \times (2.25 \times 10^{-6})

                   K = \frac{1.36 \times 10^{-4}}{2.25 \times 10^{-6}}

                      = 60.4 M^{-1}sec^{-1}

Hence, the value of rate constant for the given reaction is 60.4 M^{-1}sec^{-1} .

(c) Now, we will calculate the rate as follows.

                Rate = K [OCl^{-}][l^{-}]

                         = 60.4 \times (1.8 \times 10^{3}) \times (6.0 \times 10^{4})

                        = 6.52 \times 10^{5}

Therefore, rate when [OCl^{-}] = 1.8 \times 10^{3} M and [I^{-}]= 6.0 \times 10^{4} M is  6.52 \times 10^{5}.

8 0
3 years ago
what is the best course of action if solid material remains in the flask after the heating step of recrystallization
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Answer:

filter the hot mixture.

Explanation:

Solid is stayed undissolved since the arrangement is gotten super saturated. On the off chance that solid molecule is available recrysallization won't happen in this way we need expel the solid molecule by filtarion in hot condition itself . Subsequently, arrangement become totally homogenous and recrysallization item will shaped by moderate cooling

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3 years ago
Titration of 0.824 g of potassium hydrogen phthalate required 38.314 g of naoh solution to reach the end point detected by pheno
melamori03 [73]

1.062 mol/kg.

<em>Step 1</em>. Write the balanced equation for the neutralization.

MM = 204.22 40.00

KHC8H4O4 + NaOH → KNaC8H4O4 + H2O

<em>Step 2</em>. Calculate the moles of potassium hydrogen phthalate (KHP)

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= 4.035 mmol KHP

<em>Step 3</em>. Calculate the moles of NaOH

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= 37.973 g

<em>Step 6</em>. Calculate the molal concentration of the NaOH

<em>b</em> = moles of NaOH/kg of water = 0.040 35 mol/0.037 973 kg = 1.062 mol/kg

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