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3 years ago

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Chemistry

1 answer:

Soloha48 [4]3 years ago

7 0

Before it is released it as potential energy and after it has been released it transforms into kinetic energy.

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It is difficult to measure the correct atomic radius. Why?

Paul [167]

Answer:

its kinda you just need to simplify

Explanation:

4 0

3 years ago

commercial cold packs consist of solid NH4NO3 and water. In a coffee-cup calorimeter, 5.60g NH4NO3 is dissolved in 100g of water

fgiga [73]

Answer:

-1.37 kJ/mol

Explanation:

The expression for the calculation of the enthalpy of dissolution of [tex[NH_4NO_3[/tex] is shown below as:-

$\Delta H=m\times C\times \Delta T$

Where,

$\Delta H$ is the enthalpy of dissolution of [tex[NH_4NO_3[/tex]

m is the mass

C is the specific heat capacity

$\Delta T$ is the temperature change

Thus, given that:-

Mass of ammonium nitrate = 5.60 g

Specific heat = 4.18 J/g°C

$\Delta T=17.9-22.0\ ^0C=-4.1\ ^0C$

So,

$\Delta H=-1.25\times 4.18\times 3.9\ J=-95.9728\ J$

Negative sign signifies loss of heat.

Also, 1 J = 0.001 kJ

So,

$\Delta H=-0.096\ kJ$

Also,

Molar mass of [tex[NH_4NO_3[/tex] = 80.043 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{5.60\ g}{80.043 \ g/mol}$

$Moles= 0.06996\ mol$

Thus, $\Delta H=-\frac{0.096}{0.06996}\ kJ/mol=-1.37\ kJ/mol$

6 0

4 years ago

A 200.-milliliter sample of CO2(g) is placed in a sealed, rigid cylinder with a movable piston at 296 K and 101.3 kPa. Determine

bagirrra123 [75]

Answer:

The final volume of the sample of gas $V_{2}$ = 0.000151 $m^{3}$

Explanation:

Initial volume $V_{1}$ = 200 ml = 0.0002 $m^{3}$

Initial temperature $T_{1}$ = 296 K

Initial pressure $P_{1}$ = 101.3 K pa

Final temperature $T_{2}$ = 336 K

Final pressure $P_{2}$ = K pa

Relation between P , V & T is given by

$P_{1} \frac{V_{1} }{T_{1} } = P_{2} \frac{V_{2} }{T_{2} }$

Put all the values in the above equation we get

$101.3 (\frac{0.0002}{296} )= 152 (\frac{V_{2} }{336} )$

$V_{2}$ = 0.000151 $m^{3}$

This is the final volume of the sample of gas.

4 0

4 years ago

The radioactive isotope 210/84 Po decays by alpha emission.

yan [13]

The half-life of polonium-210, given that it decays from 98.3 micrograms to 12.3 micrograms in 414 days is 138 days

<h3>How to determine the number of half-lives </h3>

Original amount (N₀) = 98.3 micrograms
Amount remaining (N) = 12.3 micrograms
Number of half-lives (n) =?

2ⁿ = N₀ / N

2ⁿ = 98.3 / 12.3

2ⁿ = 8

2ⁿ = 2³

n = 3

<h3>How to determine the half life </h3>

Number of half-lives (n) = 3
Time (t) = 414 days
Half-life (t½) = ?

t½ = t / n

t½ = 414 / 3

t½ = 138 days

Learn more about half life:

brainly.com/question/26374513

#SPJ1

4 0

2 years ago

Read 2 more answers

As fast as possible really fast really fast

vichka [17]

$\\ \rm\longmapsto Density=\dfrac{Mass}{Volume}$

Volume=120mL
Density=0.75g/ml

$\\ \rm\longmapsto Mass=Density\times Volume$

$\\ \rm\longmapsto Mass=120(0.75)$

$\\ \rm\longmapsto Mass=90g$

7 0

3 years ago