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3 years ago

Heat would best be transferred by conduction between

2 answers:

6 0

Stove and copper pot. The candle and air transfer heat through convection, the oven cannot properly conduct heat to the glass dish, wooden spoon is a horrible conductor in any form, and the copper pot is metal so it can better conduct with the stove

lina2011 [118]3 years ago

4 0

Answer:

C. or B.

Explanation:

The reason I chose c best, is when your cooking something with a pot with food in it max heat in about 5 minutes, the handle starts to get warmer every minute. This is conduction, where the heat transfers from the very bottom all the way to the very edge of something that is heated at a certain temperature.

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The rate of disappearance of HBr in the gas phase reaction 2 HBr(g) → H2(g) + Br2(g) is 0.140 M s-1 at 150°C. The rate of appear

djverab [1.8K]

Answer: The rate of appearance of $Br_2$ is $0.0700Ms^{-1}$

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$2HBr(g)\rightarrow H_2(g)+Br(g)$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of HBr = $-\frac{1d[HBr]}{2dt}Rate in terms of appearance of [tex]H_2$ = $\frac{1d[H_2]}{dt}$

Rate in terms of appearance of $Br_2$ = $\frac{1d[Br_2]}{dt}$

$-\frac{1d[HBr]}{2dt}=\frac{d[H_2]}{dt}=\frac{d[Br_2]}{dt}$

Given :

$-\frac{1d[HBr]}{dt}=0.140Ms^{-1}$

The rate of appearance of $Br_2$ ;

$\frac{1d[Br_2]}{dt}=-\frac{1d[HBr]}{2dt}=\frac{1}{2}\times 0.140=0.0700Ms^{-1}$

Thus rate of appearance of $Br_2$ is $0.0700Ms^{-1}$

6 0

3 years ago

How many moles are in a sample of 1.52 x 1024 atoms of mercury (Hg)?

OLEGan [10]

The answer is:

2.5 moles

The explanation:

when avogadro's number = 6.02 * 10^23

and when 1 mole of the sample will give → 6.02*10^23 atoms

So, ??? mole of the sample will give→ 1.52 x 10^24 atoms

∴ number of moles = (1.52 X 10^24) * 1 mole / (6.02X10^23)

= 2.5 moles

8 0

3 years ago

Which states of matter have particles that move independently of one another with very little attraction?

Bond [772]

the answer is a- gas and plasma. i took it on usa test prep.

4 0

3 years ago

Read 2 more answers

Wie geht diese Aufgabe?

bogdanovich [222]

Answer:

1u1u2j2jj and her family

Explanation:

I have been 9in 7the 6same 6of 8for and then I am a solicitor at Cardiff on

3 0

3 years ago

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl−+I−→OI−+Cl− T

motikmotik

Answer :

(a) The rate law for the reaction is:

$\text{Rate}=k[OCl^-]^1[I^-]^1$

(b) The value of rate constant is, $60.4M^{-1}s^{-1}$

(c) rate of the reaction is $6.52\times 10^{-5}Ms^{-1}$

Explanation :

Rate law : It 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:

$OCl^-+I^-\rightarrow OI^-+Cl^-$

Rate law expression for the reaction:

$\text{Rate}=k[OCl^-]^a[I^-]^b$

where,

a = order with respect to $OCl^-$

b = order with respect to $I^-$

Expression for rate law for first observation:

$1.36\times 10^{-4}=k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b$ ....(1)

Expression for rate law for second observation:

$2.72\times 10^{-4}=k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b$ ....(2)

Expression for rate law for third observation:

$2.72\times 10^{-4}=k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b$ ....(3)

Dividing 1 from 2, we get:

$\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}\\\\2=2^a\\a=1$

Dividing 1 from 3, we get:

$\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b}\\\\2=2^b\\b=1$

Thus, the rate law becomes:

$\text{Rate}=k[OCl^-]^a[I^-]^b$

a = 1 and b = 1

$\text{Rate}=k[OCl^-]^1[I^-]^1$

Now, calculating the value of 'k' (rate constant) by using any expression.

$1.36\times 10^{-4}=k(1.5\times 10^{-3})(1.5\times 10^{-3})$

$k=60.4M^{-1}s^{-1}$

Now we have to calculate the rate for a reaction when concentration of $OCl^-$ and $I^-$ is $1.8\times 10^{-3}M$ and $6.0\times 10^{-4}M$ respectively.

$\text{Rate}=k[OCl^-][I^-]$

$\text{Rate}=(60.4M^{-1}s^{-1})\times (1.8\times 10^{-3}M)(6.0\times 10^{-4}M)$

$\text{Rate}=6.52\times 10^{-5}Ms^{-1}$

Therefore, the rate of the reaction is $6.52\times 10^{-5}Ms^{-1}$

8 0

3 years ago

Heat would best be transferred by conduction￼￼ between

Heat would best be transferred by conduction between