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1 year ago

Calcium-45 has a half-life of 162.7 days. If four half-lives have elapsed, how much time has passed?

2 answers:

7 0

Half-life<span> is the time required for a quantity to reduce to half its initial value. </span><span>If four half-lives have elapsed for calcium-45, then it would be 4x162.7 = 650.8 days have passed. Hope this answers the question. Have a nice day. Feel free to ask more questions.</span>

Len [333]1 year ago

6 0

Answer:

1.78 years

Explanation:

First half life is the time at which the concentration of the reactant reduced to half.

Second half reaction is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/4.

Third half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/8.

Fourth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/16.

There are four half lives. So, time has passed = 4*162.7 days = 650.8 days = 1.78 years

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as a 15.1-gram sample of a metal absorbs 48.75 j of heat its temperature increases 25.0 k what is the specific heat capacity of

Leviafan [203]

Specific heat capacity (c) of a material is related to the Energy Absorbed (Q), mass of the material (m) and the change in temperature (T) by the following equation:

$c= \frac{Q}{mT}$

Substituting the values of Q, m and T in the above equation, we get:

$c= \frac{48.75}{15.1*25}=0.129$

So the specific heat capacity of the metal with given conditions will be 0.129 J/g.K

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2 years ago

A sodium ion, Na+, with a charge of 1.6×10−19C and a chloride ion, Cl− , with charge of −1.6×10−19C, are separated by a distance

sasho [114]

Answer:

$W\geq 2.1x10^{-19}J$

Explanation:

Due to Coulomb´s law electric force can be described by the formula $F=K\frac{q_{1}.q_{2}}{r^{2}}$ , where K is the Coulomb´s constant ( $9x10^{9} N\frac{m^{2} }{C^{2} }$ ), $q_{1}$ = Charge 1 (Na+ in this case), $q_{2}$ is the charge 2 (Cl-) and r is the distance between both charges.

Work made by a force is W=F.d and total work produced is the change in energy between final and initial state. this is $W=W_{f} -W_{i}$ .

so we have $W=W_{f} -W_{i} =(K\frac{q_{(Na+)}q_{(Cl-)}rf}{r_{f} ^{2}})-(K\frac{q_{(Na+)}q_{(Cl-)}ri}{r_{i} ^{2}})=Kq_{(Na+)}q_{(Cl-)[\frac{1}{{r_{f}}} -\frac{1}{{r_{i}}}]$

Given that ri= 1.1nm= $1.1x10^{-9}m$ and rf= infinite distance

$W=(9x10^{9})(1.6x10^{-19})(-1.6x10^{-19})[\frac{1}{\alpha }-\frac{1}{(1.1x10^{-9})}]=2.1x10^{-19}J$

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2 years ago

A sample of ammonia has a mass of 82.9 g. how many molecules are in this sample?

alina1380 [7]

The chemical formula for ammonia is NH3. So first, you need to find the molar mass of ammonia (how many grams in one mole).
N=14g
H3=3g
So one mole of NH3 is 17 grams, you can divide 82.9 grams by 17 grams to find the number of molecules. The answer should be 4.876 moles (molecules) of ammonia. Hope this helps!

8 0

2 years ago

Read 2 more answers

Methylene chloride (CH2Cl2) has fewer chlorine atoms than chloroform (CHCl3). Nevertheless, methylene chloride has a larger mole

icang [17]

Answer:

Explanation has been given below.

Explanation:

Chloroform has three polar C-Cl bonds. Methylene chloride has two polar C-Cl bonds. So it is expected that chloroform should be more polar and posses higher dipole moment than methylene chloride.
Two factors are liable for the opposite trend observed in dipole moments of methylene chloride and chloroform.
First one is the number of hyperconjugative hydrogen atoms present in a molecule. Hyperconjugation occurs with vacant d-orbital of Cl atom. Hyperconjugation amplifies charge separation in a molecule resulting higher dipole moment.
Methylene chloride has two hyperconjugative hydrogen atoms and chloroform has one hyperconjugative hydrogen atom.Therefore methylene chloride should have higher charge separation as compared to chloroform.
Second one is induction of opposite polarity in a C-Cl bond by another C-Cl bond in a molecule. Higher the opposite induction of polarity, lower the charge separation in a molecule and hence lower the dipole moment of a molecule.
Chloroform has three C-Cl bonds and methylene chloride has two C-Cl bonds. Therefore opposite induction is higher for chloroform resulting it's lower dipole moment.