575 394 9550<br> bFzN0S<br> come study with me!

What is the size comparison between earth & mars

Neko [114]

Answer:

earth has a size of 6,371 km wich is greater than mars size 3,389.5 km

Explanation:

3 0

3 years ago

When 2.69 g 2.69 g of a nonelectrolyte solute is dissolved in water to make 345 mL 345 mL of solution at 26 °C, 26 °C, the solut

Gre4nikov [31]

Answer:

The molar concentration of this solution is 0.0463 mol/L

Explanation:

Step 1 : Data given

Mass of a nonelectrolyte solute = 2.69 grams

Volume of water = 345 mL = 0.345 L

Temperature = 26.0°CC = 273 + 26 = 299 K

The osmotic pressure = 863 torr

⇒ 863torr /760 = 1.13553 atm

Step 2: Calculate the molar concentration of this solution

Π = i*M*R*T

⇒with Π = the osmotic pressure = 1.13553 atm

⇒with i = the van't Hoff factor of the nonelectrolyte solute = 1

⇒with M = the molar concentration = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 299 K

1.13553 atm = 1 * M * 0.08206 L*atm/mol*K * 299 K

M = 1.13553 / (0.08206*299)

M = 0.0463 mol/L

The molar concentration of this solution is 0.0463 mol/L

5 0

3 years ago

IF ANSWER I WILL GIVE BRAINLESS

vovikov84 [41]

Answer:

Answer 9 - 100 joules energy was at the producer level

Answer 10 - Remaining energy is used in metabolism

Explanation:

Answer 9

The energy at each trophic level is only 10% of the energy at its previous trophic level.

The energy at producer level is X

$10$ % of $X = 10$ Joules

$\frac{10}{100} * X = 10\\X = 100$ Joules

Answer 10

Because the remaining 90% energy is utilized by the producer for its metabolism

3 0

3 years ago

Classify each process as either physical or a chemical change.

Dafna11 [192]

Answer:

I think they are all correct

7 0

3 years ago

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

575 394 9550 bFzN0S come study with me!