What is the study of organism at the early stages of devolopment

Alekssandra [29.7K]

Answer:

<h3>Sand cannot mixed on water not float on water .</h3>

<h3>When we mix sand and water , No reaction take place . The sand simply settles down at the bottom of the water container . This is why sand is heavier than water and therefore cannot float in water .</h3>

<h2>Hope this helps you ✌️</h2>

7 0

2 years ago

Consider the following reaction at equilibrium: 2NH3 (g) N2 (g) + 3H2 (g) Le Châtelier's principle predicts that the moles of H2

Lilit [14]

Answer:

A decrease in the total volume of the reaction vessel (T constant)

Explanation:

Le Châtelier's principle predicts that the moles of H2 in the reaction container will increase with a decrease in the total volume of the reaction vessel.
<em><u>According to the Le Chatelier's principle, when a chnage is a applied to a system at equilibrium, then the equilibrium will shift in a way that counteracts the effect causing it.</u></em>
In this case, a decrease in volume means there is an increase in pressure, therefore the equilibrium will shift towards the side with the fewer number of moles of gas.

3 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

Someone please help i don’t have much time left

REY [17]

Answer: Energy of reactants = 30, Energy of products = 10

Exothermic

Activation energy for forward reaction is 10.

Explanation:

Exothermic reactions are defined as the reactions in which energy of the product is lesser than the energy of the reactants. The total energy is released in the form of heat and $\Delta H$ for the reaction comes out to be negative.