Plz help me with this question

Which one would it be <br> 1,2,3,4

sesenic [268]

I think it's 1. PCl5= PCl3 + Cl2. I can't tell the exact reason but it's something to do with Atomic Redox Reaction

6 0

3 years ago

Biochemical reactions in your cells occur best at approximately which temperature

VARVARA [1.3K]

Answer: 98.6˚F

Explanation:

3 0

3 years ago

Select the most likely product for this reaction: upper a g upper n upper o subscript 3 (a q) plus upper n a upper o upper h (a

Mamont248 [21]

One of the most likely products for the reaction would be $Ag_2O$

<h3>Chemical reactions</h3>

The reaction between $AgNO_3$ and $NaOH$ yields 3 products which are $Ag_2O$ (a precipitate), $NaNO_3$ , and $H_2O$ as shown by the equation below:

$2 AgNO_3 + 2 NaOH --- > Ag_2O + 2 NaNO_3 + H_2O$

One of the products precipitates out of the solution to give the reaction a precipitation reaction look.

More on precipitation reaction can be found here: brainly.com/question/24158764

#SPJ4

5 0

2 years ago

Read 2 more answers

Freeze-drying is a process used to preserve food. If strawberries are to be freeze-dried, then they would be frozen to -80.00 °C

katrin2010 [14]

Answer:

1. 389 kJ; 2. 7.5 µg; 3. 6.25 days

Explanation:

1. Energy required

The water is converted directly from a solid to a gas (sublimation).

They don't give us the enthalpy of sublimation, but

$\Delta_{\text{sub}}H = \Delta_{\text{fus}}H + \Delta_{\text{vap}}H = 6.01 + 40.68 = 46.69 \text{ kJ}\cdot\text{mol}^{-1}$

The equation for the process is then

Mᵣ: 18.02

46.69 kJ + H₂O(s) ⟶ H₂O(g)

m/g: 150

(a) Moles of water

$\text{Moles} = \text{150 g} \times \dfrac{\text{1 mol}}{\text{18.02 g}} = \text{8.324 mol}$

(b) Heat removed

46.69 kJ will remove 1 mol of ice.

$\text{Heat removed} = \text{8.234 mol} \times \dfrac{\text{46.69 kJ}}{\text{1 mol}} = \textbf{389 kJ}\\\text{It takes $\large \boxed{\textbf{389 kJ}}$ to remove 150 g of ice}$

2. Mass of water vapour in the freezer

For this calculation, we can use the Ideal Gas Law — pV = nRT

(a) Moles of water

Data:

$p = 1.00 \times 10^{-3}\text{ torr } \times \dfrac{\text{1 atm}}{\text{760 torr}} = 1.316 \times 10^{-6}\text{ atm}$

V = 5 L

T = (-80 + 273.15) K = 193.15 K

Calculation:

$\begin{array}{rcl}pV & = & nRT\\1.316 \times 10^{-6}\text{ atm} \times \text{5 L} & = & n \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{193.15 K }\\6.6 \times 10^{-6} & = & 15.85n\text{ mol}^{-1} \\n & = & \dfrac{6.6 \times 10^{-6}}{15.85\text{ mol}^{-1}}\\\\& = & 4.2 \times 10^{-7} \text{ mol}\\\end{array}$

(b) Mass of water

$\text{Mass} = 4.2 \times 10^{-7} \text{ mol} \times \dfrac{\text{18.02 g}}{\text{1 mol}} = 7.5 \times 10^{-6}\text{ g} = 7.5 \, \mu \text{g}\\\\\text{At any given time, there are $\large \boxed{\textbf{7.5 $\mu$g}}$ of water vapour in the freezer.}$