How to test for sugar in your gingerbread

Need help ASAP with this question

Karo-lina-s [1.5K]

Answer: noble gasses

Explanation: zoom on on the red section and look at the key for what red means.

8 0

3 years ago

A solution of sodium chloride in water has a vapor pressure of 19.6 torr at 25°C. What is the mole fraction of NaCl in this solu

Debora [2.8K]

Answer:

Mole fraction of Nacl is 0.173

Explanation:

we know that

$P_{sol}=\chi_{solvent}P^0_{solvent}$

where,

P sol - the vapor pressure of the solution

χ solvent - the mole fraction of the solvent

P ∘ solvent - the vapor pressure of the pure solvent

This means that in order to be able to calculate the mole fraction of sodium chloride, you need to know what the vapor pressure of pure water is at

25 ° C You can use an online calculator to find that the vapor pressure of pure water at 25 C is equal to about 23.8 torr .

$\chi_{water}= \frac{P{sol}}{P^0{water}}$

$\chi_{water}= \frac{19.6}{23.8}$

=0.827

Also we know that

$\chi_{water}+\chi_{Nacl}= 1$

This means that the mole fraction of sodium chloride is

χ_{Nacl}= 1-Χ_{water}

= 1-0.827 =0.173

3 0

3 years ago

In a room full of air, the air is mainly composed of Nitrogen and Oxygen molecules (both at room temperature). Find (to two sign

zloy xaker [14]

Explanation:

Expression for the $v_{rms}$ speed is as follows.

$v_{rms} = \sqrt{\frac{3kT}{M}}$

where, $v_{rms}$ = root mean square speed

k = Boltzmann constant

T = temperature

M = molecular mass

As the molecular weight of oxygen is 0.031 kg/mol and R = 8.314 J/mol K. Hence, we will calculate the value of $v_{rms}$ as follows.

$v_{rms} = \sqrt{\frac{3kT}{M}}$

= $\sqrt{\frac{3 \times 8.314 J/mol K \times 309.02 K}{0.031 kg/mol}}$

= 498.5 m/s

Hence, $v_{rms}$ for oxygen atom is 498.5 m/s.

For nitrogen atom, the molecular weight is 0.028 kg/mol. Hence, we will calculate its $v_{rms}$ speed as follows.

$v_{rms} = \sqrt{\frac{3kT}{M}}$

= $\sqrt{\frac{3 \times 8.314 J/mol K \times 309.92 K}{0.028 kg/mol}}$

= 524.5 m/s

Therefore, $v_{rms}$ speed for nitrogen is 524.5 m/s.

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

How many moles of electrons must be transferred to plate out 110 g of manganese (MW ~ 55g/mol) from a solution of permanganate (

ololo11 [35]

Answer:

14 mol e⁻

Explanation:

Step 1: Write the balanced half-reaction for the reduction of permanganate to manganese

8 H⁺(aq) + 7 e⁻ + MnO₄⁻(aq) ⇒ Mn(s) + 4 H₂O(l)

Step 2: Calculate the moles corresponding to 110 g of manganese

The molar mass of Mn is 55 g/mol.

110 g × 1 mol/55 g = 2 mol

Step 3: Calculate the number of moles of electrons needed to produce 2 moles of Mn

According to the half-reaction, 7 moles of electrons are required to produce 1 mole of Mn.

2 mol Mn × 7 mol e⁻/1 mol Mn = 14 mol e⁻

8 0

3 years ago