Ask question

Ask question

All categories

3 years ago

8

Which of the following produces a chemical change?

1 answer:

Aleksandr [31]3 years ago

8 0

Answer:

B. Grilling a piece of meat

since the process involves burning

You might be interested in

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

3 years ago

Which choice is not true of a liquid in a glass capillary with a convex meniscus?

Mamont248 [21]

I believe <span>The liquid has strong cohesive forces.</span>

5 0

4 years ago

Read 2 more answers

Calculate the RF value - Distance travelled by substance x=5cm Distance travelled by solvent = 13cm

Anna35 [415]

Answer:

The rf value is 0.38

Explanation:

You do 5/13 which gives the answer 0.38

7 0

3 years ago

crimeas [40]

Answer:

because.......

...........

.........

3 0

3 years ago

0.456 grams of a monoprotic acid is titrated with 45.88 mL of 0.0500 M NaOH. What is the molecular mass (molar mass) of the acid

aivan3 [116]

Answer:

$Molar\ mass= 198.78\ g/mol$

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For NaOH :

Molarity = 0.0500 M

Volume = 45.88 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 45.88×10⁻³ L

$Moles_{NaOH} =0.0500 \times {45.88\times 10^{-3}}\ moles=0.002294\ moles$

Moles of NaOH = Moles of monoprotic acid

So, moles of monoprotic acid = 0.002294 moles

Given that:- mass = 0.456 g

The formula for the calculation of moles is shown below:

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

Thus,

$0.002294\ moles= \frac{0.456\ g}{Molar\ mass}$

$Molar\ mass= 198.78\ g/mol$

3 0

3 years ago