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umka2103 [35]
3 years ago
7

What is the molar mass for Agno3​

Chemistry
1 answer:
Aloiza [94]3 years ago
5 0

Answer:

169.87 g/mol

Explanation:

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Which daughter element is produced from the beta decay of 89/36 kr
Vesnalui [34]

Answer:

89 37 Rb

Explanation:

just the facts

6 0
4 years ago
What temperature in kelvin does 60.5 liters of sulfur dioxide occupy if there are 2.5 mol at 0.75 atm??
7nadin3 [17]
 The temperature  in kelvin  does  60.5  liters of sulfur  dioxide  occupy  if there  are  2.5  mole  at 0.75 atm  is   221.07  kelvin 

              Explanation
      This is  calculated using   ideal  gas  equation, that is  PV=nRT
where,  P(pressure)  = 0.75 atm
             V(volume)  = 60.5 L
             n(moles)  = 2.5 mole
             R( gas  constant) = 0.0821 L.atm/mol.k
            T(temperature  =?

by  making   T  the subject   of  the formula  
T  is  therefore =Pv/nR

T= (0.75 atm  x  60.5 L) / (  2.5 molex 0.0821  L.atm/mol.K)  = 221.07 kelvins

6 0
3 years ago
Many double-displacement reactions are enzyme-catalyzed via the "ping pong" mechanism, so called because the reactants appear to
zhenek [66]

Answer:

<u>D. It will decrease by a factor of 4</u>

Explanation:

According to the question , the equation follows :

A+B\rightarrow C+D

Rate law : This states the rate of reaction is directly proportional to concentration of reactants with each reactant raised to some power which may or may not be equal to the stoichiometeric coefficient.

Rate\ \alpha [A]^{a}[B]^{b}

r=[A]^{a}[B]^{b}.................(1)

STEP": First, find out the power "a" and "b"

a+b = 3 (because it is given that the reaction follow 3rd order-kinetics)

According to question, <u><em>doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,</em></u>

r' = 2r if [A'] = 2[A]

Here [B] is uneffected means [B']=[B]

hence new rate =

r'=[A']^{a}[B']^{b}

Put the value of [A'] , [B'] and r' in the above equation:

2r=[2A]^{a}[B]^{b}...........(2)

Divide equation (1) by (2) we , get

\frac{2r}{r}=\frac{[2A]^{2}[B]^{b}}{[A]^{a}[B]^{b}}

2= 2(\frac{A}{A})^{a}\times (\frac{B}{B})^{b}

Here A and A cancel each other

B and B cancel each other

We get,

2= 2^{a}\times 1^{b}

1^b = 1 ( power of 1 = 1)

2= 2^{a}

This is possible only when a = 1

We know that : a + b = 3

1 + b = 3

b =3 -1  = 2

b = 2

Hence the rate law becomes :

r=[A]^{a}[B]^{b}

<u>r=[A]^{1}[B]^{2}.............(3)</u>

Look in the question now, it is asked to calculate the concentration of [B],if  cut in half

Hence

[B']=1/2[B]

Insert the value of [B'] in equation (3)

r'=[A]^{1}[B']^{2}

r'=[A]^{1}(\frac{1}{2}[B])^{2}

r'=\frac{1}{4}[A]^{1}[B]^{2}............(a)

But

r=[A]^{a}[B]^{b}..............(b)

Compare equation (a) and (b) , we get

new rate r' =

<u>r' = 1/4 r</u>

7 0
3 years ago
An anode is an electrode immersed in the half-cell where _____ takes place.
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If I'm not mistaken, the answer should be B. Oxidation
5 0
4 years ago
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The correctness of a scientific claim comes from
likoan [24]
It's C. but come on, use some common sense.
8 0
4 years ago
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