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Sholpan [36]
3 years ago
10

Sometimes in lab we collect the gas formed by a chemical reaction over water (see sketch at right). This makes it easy to isolat

e and measure the amount of gas produced.
Suppose the CO gas evolved by a certain chemical reaction taking place at 55 degree C is collected over water, using an apparatus something like that in the sketch, and the final volume of gas in the collection tube is measured to be 117 ml.

Calculate the mass of CO that is in the collection tube. Round your answer to 2 significant digits. You can make any normal and reasonable assumption about the reaction conditions and the nature of the gases.
Chemistry
1 answer:
zepelin [54]3 years ago
3 0

Answer:

The correct answer is 0.12 grams.

Explanation:

The mass of carbon monoxide or CO collected in the tube can be determined by using the ideal gas equation, that is, PV = nRT.

Based on the given question, P or the pressure of the gas is given as 1 atm, volume of the gas collected in the tube is 117 ml or 0.117 L.  

The number of moles or n can be determined by using the equation, mass/molar mass.  

R is the universal gas constant, whose value is 0.0821 L atmK^-1mol^-1, and temperature is 55 degree C or 328 K (55+273).  

On putting the values we get:

n = PV/RT

= (1 atm*0.117 L) / (0.0821 L atmK^-1mol^-1 * 328 K)

= 0.0043447 mol

Therefore, mass of CO will be moles * molar mass of CO

= 0.0043447 mol * 28 g/mol

= 0.12 g

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Answer:

1. Reflex Conditioning

Explanation:

Conditioning is an aspect of learning where a stimulus works effectively in producing a response from an organism. This response becomes regular given the type of reinforcement that is administered to the organism. The reinforcement is usually a reward that is given to the organism.

This is what is observed in the woodpecker.  The stimulus or reinforcement which proves effective in producing a continuous response  from the woodpecker is the abundant supply of the birds favorite bugs. This reinforcement makes it possible for the woodpecker to become conditioned towards returning to the tree.

4 0
3 years ago
Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f
N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+2B\rightleftharpoons C

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b

where,

a = order with respect to A

b = order with respect to B

  • Expression for rate law for first trial:

5.4\times 10^{-3}=k(0.30)^a(0.050)^b ....(1)

  • Expression for rate law for second trial:

1.1\times 10^{-2}=k(0.30)^a(0.100)^b ....(2)

  • Expression for rate law for third trial:

2.2\times 10^{-2}=k(0.50)^a(0.050)^b ....(3)

Dividing 2 by 1, we get:

\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1

Dividing 3 by 1, we get:

\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^1       ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}

Calculating the initial rate of formation of C by using equation 4, we get:

k=1.2M^{-2}s^{-1}

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}

Hence, the initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

8 0
3 years ago
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Answer:Molecules

Explanation:

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Answer is: because alkaline metals (group IA metals) are the strongest reducing agents and most reactive metals.
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Alkaline metals tend to lose only one electron in redox reaction.</span>
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All human body systems works together m. What would happen to someone if one of their systems becomes inefficient?
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2 years ago
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