2 Na + 1 Cl2 → 2 Naci<br> How many grams of NaCl are created from 23,2 grams Cl2?

Indigenous rocks form by the cooling and hardening of melted rock.* True False

salantis [7]

Answer: True

Explanation: They are formed by the cooling and hardening of molten magma

3 0

3 years ago

Jason ran 5/7 of the distance around the school track sara ran 4/5 of Jason distance what fraction of the total distance around

FrozenT [24]

Let x represent the total distance around the track
Jason's distance: (5/7)x
Sara ran (4/5) of Jason's distance,
so she ran (4/5)*(5/7)x = (4/7)x
Sara ran 4/7 of the total distance

7 0

3 years ago

A sample of gas occupies 280 mL when the pressure is 560.00 mm Hg . If the temperature remains constant , what is the new pressu

vichka [17]

Answer : The new pressure if the volume changes to 560.0 mL is, 280 mmHg

Explanation :

According to the Boyle's, law, the pressure of the gas is inversely proportional to the volume of gas at constant temperature and moles of gas.

$P\propto \frac{1}{V}$

or,

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure = 560.00 mmHg

$P_2$ = final pressure = ?

$V_1$ = initial volume = 280 mL

$V_2$ = final volume = 560.0 mL

Now put all the given values in the above formula, we get:

$560.00mmHg\times 280 mL=P_2\times 560.0 mL$

$P_2=280mmHg$

Therefore, the new pressure if the volume changes to 560.0 mL is, 280 mmHg

3 0

4 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

3 years ago

Question 11. Identify the reducing agent <br> Sn+2 + AG 0 —> Sn0 + Ag+

Temka [501]

Answer:

Ag 0 is the reducing agent.

Explanation:

Reducing -> gaining electrons

Oxidizing -> losing electrons

Ag lost electrons (became more positive) since it went from a 0 charge to a +1 charge. Therefore it was oxidized. Ag+ is the oxidized product. Reactants that create an oxidized product are called reducing agents. This would make Ag 0 the reducing agent in this reaction.

4 0

3 years ago

2 Na + 1 Cl2 → 2 Naci How many grams of NaCl are created from 23,2 grams Cl2?