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3 years ago

Part iii. Is the dissolution reaction of ammonium nitrate enthalpy favored?

Chemistry

1 answer:

lesya [120]3 years ago

3 0

Answer:

No, it is not enthalpy favored since the chemical system gains energy.

Explanation:

The dissolution of ammonium nitrate in water is an endothermic process.

Endothermic process requires the system to gain energy to can dissolve the particles in water.

So, the reaction is not enthalpy favored.

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All cells _____.

xeze [42]

The answer is the fourth choice, or Have membrane-bound organelles. Hope this helps you!

Please mark as Brainliest if this was correct!

-Belle

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3 years ago

The freezing point of diet soda is higher than the freezing point of regular soda, but lower than 0 degrees celcius, the freezin

Zepler [3.9K]

Answer:

Explanation:

Both Diet Soda and regular soda contain sweeteners.

When a solute is dissolved in solution, the solution undergoes *freezing point depression* it freezing point reduces. The magnitude of freezing point depression is directly proportional to the amount of solute in a solution.

Since soda Both regular or diet soda contains more solute than water , their freezing point is will consequently be lower than water

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3 years ago

BRAINLIESTTT ASAP!!! PLEASE ANSWER!

sladkih [1.3K]

If the item or substance has changed into a state where you are able to change it back such as:

Water melting

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3 years ago

Read 2 more answers

If you are calculating the grams to mL ratio, you are trying to find the object’s…

Alex

Answer:

Density

Explanation:

The ratio of mass to the volume of an object is called its density. Unit of mass is grams and that of volume is mL.

Density = mass/volume

$\text{Density}=\dfrac{\text{grams}}{\text{mL}}\\\\=\text{grams/mL}$

If you are calculating the grams to mL ratio, it means that we are trying to find the object's density.

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3 years ago

calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w

kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is $-22Jmole^-^1k^-^1$

(b) The normal boiling point of water (J·K−1·mol−1) is $-109Jmole^-^1K^-^1$

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is 109J/mole

Explanation:

Lets calculate

(a) - General equation -

$(\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p$ = $-5_m(\beta )+5_m(\alpha )$ = $-\frac{\Delta H}{T}$

$\alpha ,\beta$ → phases

ΔH → enthalpy of transition

T → temperature transition

$(\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p$ = $= -\frac{\Delta_fH}{T_f}$

= $\frac{-6.008kJ/mole}{273.15K}$ ( $\Delta_fH$ is the enthalpy of fusion of water)

= $-22Jmole^-^1k^-^1$

(b) $(\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}$

= $\frac{40.656kJ/mole}{373.15K}$ ( $\Delta_v_a_p_o_u_rH$ is the enthalpy of vaporization)

= $-109Jmole^-^1K^-^1$

$[\mu(l-5$ ° $C)-\mu(l,0$ ° $C)]$ = $[\mu(s-5$ ° $C)-\mu(s,0$ ° $C)]$ $=-S_m$ ΔT

$\mu(l,-5$ ° $C)-\mu(s,-5$ ° $C)=-Sm\DeltaT [\mu(l,0$

$\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)$

= 109J/mole

6 0

2 years ago