The diagram shows a model of an atom. Who first proposed this model?

Two solutions, initially at 24.60 °C, are mixed in a coffee cup calorimeter (Ccal = 15.5 J/°C). When a 100.0 mL volume of 0.100

yulyashka [42]

Answer:

ΔH = -59.6kJ/mol

Explanation:

The reaction that occurs between Ag⁺ and Cl⁻ ions is:

Ag⁺ + Cl⁻ → AgCl(s) + ΔH

To find ΔH we need to obtain moles of reaction and heat released in the reaction because ΔH is defined as heat released per mole of reaction.

Moles of reaction:

Moles of Ag⁺ and Cl⁻ added are:

Ag⁺: 0.100L * (0.100mol / L) = 0.01moles

Cl⁻: 0.100L * (0.200mol / L) 0 0.02 moles

That means limiting reactant is Ag⁺ and moles of reaction are 0.01 moles

Heat released:

To find heat released we must use coffe cup calorimeter equation:

Q = C*m*ΔT

Where C is specific heat of solution (4.18J/g°C), m is the mass of solution (200g because there are 100 + 100mL = 200mL and density of solution is 1g/mL) and ΔT is change in temperature (25.30°C - 24.60°C = 0.70°C).

Replacing:

Q = C*m*ΔT

Q = 4.18J/g°C * 200g * 0.70°C

Q = 585,2J

Is total heat released.

The calorimeter absorbs:

15.5J / °C * 0.7°C = 10.85

Thus, when 0.01 moles reacts, 585.2J + 10.85 = 596.05J are released (Heat released is heat abosrbed by calorimeter + Heat absorbed by water) and ΔH is:

ΔH = 596.05J / 0.01 moles =

ΔH = 59605J / mol =

As heat is released, ΔH < 0.

6 0

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)

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

(c) $\Delta\mu =\Delta\mu(l)-\Delta\mu(s)$ = $-S_m\DeltaT$

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

Will sodium iodide react with bromine to produce sodium bromide and iodine? Why or why not?

Gre4nikov [31]

Answer:

C

Explanation:

The higher the period the higher the activity of an element, therefore, since iodine is in period 6 and bromine is in period 5, the described reaction is not possible due to the fact that bromine is less active

4 0

2 years ago

In the sn2 experiment, what was the purpose of washing the distilled product with 5% naoh

photoshop1234 [79]

The purpose of washing the product with NaOH is simply to neutralize any acid which remained or leaked after the 1st initial separation. The NaOH base reacts with the acid to form neutralization reaction products which are soluble in water.

5 0

3 years ago

Please help!

Dahasolnce [82]

Answer:

reproduction

Explanation:

reproduction, process by which organisms replicate themselves

6 0

3 years ago