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3 years ago

44.2124+0.81+7.335 significant digits

Chemistry

1 answer:

zhenek [66]3 years ago

4 0

Explanation:

We need to understand and apply the rules of significant figures when carrying out addition operations to solve this problem:

Rules of significant figures:

Non-zero digits are always significant in a given number
Any zeros between two significant digits are significant.
A final zero or trailing zeros in the decimal portion is significant.
Any leading zero in a decimal is significant.

In addition, the answer of the sum is given to the least number of significant digits of the given numbers:

44.2124

+ 0.81

+ 7.335

sum 52.3574

Now the least number of significant figure is 2 as seen in 0.81

we round the sum up to 52

learn more:

Significant numbers brainly.com/question/6384538

#learnwithBrainly

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Given the E0 values of the following two half-reactions: Zn  Zn2+ + 2e- E0 = 0.763 volt Fe  Fe2+ + 2e- E0 = 0.441 volt a) Writ

34kurt

Answer:

For a: The balanced chemical equation is written below.

For b: The corrosion of iron pipe will take place in the presence of zinc.

For c: Zinc will not protect iron pipe from corrosion.

Explanation:

For a:

The given half reaction follows:

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, zinc will undergo reduction reaction will get reduced.

Oxidation half reaction: $Zn\rightarrow Zn^{2+}+2e^-;E^o_{Zn^{2+}/Zn}=-0.763V$

Reduction half reaction: $Fe+2e^-\rightarrow Fe;E^o_{Fe^{2+}/Fe}=-0.441V$

The balanced chemical equation follows:

$Fe+Zn^{2+}\rightarrow Fe^{2+}+Zn$

For b:

For a reaction to be spontaneous (thermodynamically feasible) , the standard electrode potential must be positive.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Calculating the $E^o_{cell}$ using above equation, we get:

$E^o_{cell}=-0.441-(-0.763)=0.322V$

As, the EMF is coming out to be positive, the reaction will be thermodynamically feasible and corrosion of iron pipe will take place in the presence of zinc.

For c:

As, the EMF of the cell is positive, the zinc will not protect the iron pipe from corrosion and the reaction will take place.

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The rate law of the reaction NH3 + HOCl → NH2Cl + H2O is rate = k[NH3][HOCl] with k = 5.1 × 106 L/mol·s at 25°C. The reaction is

SIZIF [17.4K]

Answer:

40% of the ammonia will take 4.97x10^-5 s to react.

Explanation:

The rate is equal to:

R = k*[NH3]*[HOCl] = 5.1x10^6 * [NH3] * 2x10^-3 = 10200 s^-1 * [NH3]

R = k´ * [NH3]

k´ = 10200 s^-1

Because k´ is the psuedo first-order rate constant, we have the following:

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b/(b-x) = 1.67

log(b/(b-x)) = log(1.67)

log(b/(b-x)) = 0.22

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