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3 years ago

How many seconds are in one year? how many seconds are in 1 year

Chemistry

1 answer:

Alja [10]3 years ago

5 0

Answer:

Hey mate.....

Explanation:

This is ur answer....

<em>31,536,000 seconds </em>

<em>drum roll, please… one year would equal 365 times 24 times 60 times 60 seconds…or 31,536,000 seconds! That's over 31 million seconds you have to spend over the next year....</em>

Hope it helps u,

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When discussing solutions, the amount of solute dissolved in the solvent is important to know

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Answer: B

Explanation:

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Which of the following belong to a category called the main group of elements

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Answer:

C. Alkaline Metal ok thanks

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3 years ago

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What is the half-life of an isotope that decays to 12.5% of its original activity in 19.8 hours?

katen-ka-za [31]

$x= x_{0} e^{kt} ,where~ x ~is~ the ~ |amount\ of \material\ at\ any~~time~t
$
$and~ x_{0} ~the~original~amount.$
$when~x= \frac{12.5}{100} x_{0} }= \frac{125}{1000} x_{0}$
$\frac{x}{ x_{0} } = \frac{125}{1000}= \frac{1}{8}$
$when t=19.8 hrs,
x= x_{0} e^{19.8 k}, 
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ln \frac{1}{8} =19.8 k,
$
$ln1-ln8=19.8k,
0-ln 2^{3} =19.8 k,
-3 ln2=19.8k
k= \frac{-3 ln2}{19.8} = \frac{- ln 2}{6.6}$ $x= x_{0} e^{ \frac{-ln2}{6.6}t } ,
 \frac{x}{ x_{0} } = e^{ \frac{- ln 2}{6.6}t } 
$ $\frac{1}{2} = e^{ \frac{- ln2}{6.6}t } = \frac{1}{ e^{ \frac{ln 2}{6.6} t} } ,
2= e^{ \frac{ln2}{6.6}t } ,
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8 0

4 years ago

What is the standard potential, e∘celle∘cell, for this galvanic cell? use the given standard reduction potentials in your calcul

Olin [163]

The standard potential for the given galvanic cell is 0.477 V

<h3>What is electrode potential?</h3>

The electrode potential is the electromotive force of a galvanic cell built using a standard reference electrode and another electrode whose potential is to be found.

There are two types of electrode potential

Oxidation potential - The potential associated with oxidation reaction is known as oxidation potential

Reduction potential - The potential associated with reduction reaction is known as reduction potential

At the anode, oxidation occurs

$Sn(s)\rightarrow Sn^{2+}(aq)+2e^-$

At the cathode, reduction occurs

$Cu^{2+}(aq)+2e^-\rightarrow Cu(s)$

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

= 0.337 - (-0.140)

= 0.477 V

Thus, The standard potential for the given galvanic cell is 0.477 V

Learn more about electrode potential:

brainly.com/question/17362810

#SPJ4

Disclaimer: The question was given incomplete on the portal. Here is the complete question

Question: What is the standard potential, E∘cell, for this galvanic cell? Use the given standard reduction potentials in your calculation as appropriate.

$Sn^{2+}(aq)+2e^-\rightarrow Sn(s)$ , E°red=−0.140 V

$Cu^{2+}(aq)+2e^-\rightarrow Cu(s)$ , E°red=+0.337 V

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2 years ago

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Select all that apply. Which of the following bases are strong enough to deprotonate CH3CH2CH2Câ¡CH (pKa = 25), so that equilibr

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Answer:

b, c, and d.

Explanation:

In an acid-base reaction, the equilibrium favors the weaker acid, thus, to the reaction favors it, the product must be weaker than the acid given. As higher is the value of pKa, as weak is the acid. Thus, the base must form a conjugate acid with pKa > 25.

The values of pKa can be found in equilibrium tables. The salts will dissociate to form the acids, which will gain one proton (H+)

a. CH3ONa -> Conjugate acid CH3OH, pKa = 15.5

b. NaNH2 -> Conjugate acid NH3, pKa = 38

c. NaH -> Conjugate acid H2, pKa = 35

d. CH3CH2Li -> Conjugate acid CH3CH3, pKa = 50

e. H2O -> Conjugate acid H3O+, pKa = -1.7

f. NH3 -> Conjugate acid NH4+, pKa = 9.2

Thus, the ones with pKa > 25 are NH3, H2, and CH3CH3, thus the bases are NaNH2, NaH, and CH3CH2Li.

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3 years ago