At which temperature will the least amount of time pass before someone detects the odor of a fragrant flower?

What is a trade-off? Don't mind the highlighted yellow circle

kirill [66]

Answer:

a new scientific discovery that benefits the environment

5 0

2 years ago

Anna71 [15]

Na 1s²2s²2p⁶3s¹
↓ - e⁻
Na⁺ 1s²2s²2p⁶ 2+2+6=10 e⁻

10 electrons are in sodium ion Na⁺

4 0

3 years ago

Which of the following is an example of a reaction that you've seen in your everyday life ?

soldier1979 [14.2K]

I believe this pertains to chemical reaction? So my guess would be answer A “Fire burning”

Don’t quote me though lol.

7 0

2 years ago

Read 2 more answers

What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )

yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution : Given,

Concentration (c) = 0.150 M

Acid dissociation constant = $k_a=4.5\times 10^{-4}$

The equilibrium reaction for dissociation of $HNO_2$ (weak acid) is,

$HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+$

initially conc. c 0 0

At eqm. $c(1-\alpha)$ $c\alpha$ $c\alpha$

First we have to calculate the concentration of value of dissociation constant $(\alpha )$ .

Formula used :

$k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}$

Now put all the given values in this formula ,we get the value of dissociation constant $(\alpha )$ .

$4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}$

$4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2$

$0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0$

By solving the terms, we get

$\alpha=0.0533$

No we have to calculate the concentration of hydronium ion or hydrogen ion.

$[H^+]=c\alpha=0.150\times 0.0533=0.007995 M$

Now we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (0.007995 M)$

$pH=2.097\approx 2.1$

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

4 0

3 years ago

A calorimeter contains 22.0 mL of water at 14.0 ∘C . When 2.50 g of X (a substance with a molar mass of 82.0 g/mol ) is added, i

Alik [6]

Answer:

The enthalpy change in the the reaction is -47.014 kJ/mol.

Explanation:

$X(s)+H_2O(l)\rightarrow X(aq)$

Volume of water in calorimeter = 22.0 mL

Density of water = 1.00 g/mL

Mass of the water in calorimeter = m

$m=1.00 g/mL\times 22.0 mL=22 g$

Mass of substance X = 2.50 g

Mass of the solution = M = 2.50 g + 22 g = 24.50 g

Heat released during the reaction be Q

Change in temperature =ΔT = 28.0°C - 14.0°C = 14.0°C

Specific heat of the solution is equal to that of water :

c = 4.18J/(g°C)

$Q=Mc\times \Delta T$

$Q=24.50 g\times 4.18 J/g ^oC\times 14.0^oC=1,433.74 J=1.433 kJ$

Heat released during the reaction is equal to the heat absorbed by the water or solution.

Heat released during the reaction =-1.433 kJ

Moles of substance X= $\frac{2.50 g}{82.0 g/mol}=0.03048 mol$

The enthalpy change, ΔH, for this reaction per mole of X:

$\Delta H=\frac{-1.433 kJ}{0.03048 mol}=-47.014 kJ/mol$

5 0

2 years ago