How do i <br> Round the value 44.981 g to three significant figures.

A solution at 25 ∘C has pOH = 10.53. Which of the following statements is or are true? The solution is acidic. The pH of the sol

Lady bird [3.3K]

Answer:

The true statements are:

The solution is acidic

The pH of the solution is 14.00 - 10.53.

$10^{-10.53}=[OH^-]$

Explanation:

The pH of the solution is defined as negative logarithm of hydrogen ion concentration present in the solution .

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

The pH value more 7 means that hydrogen ion concentration is less ,alkaline will be the solution.
The pH value less 7 means that hydrogen ion concentration is more ,acidic will be the solution.
The pH value equal to 7 indicates that the solution is neutral.

The pOH of the solution is defined as negative logarithm of hydroxide ion concentration present in the solution .

$pOH=-\log[OH^-]$

The pOH of the solution = 10.53

$10.53=-\log[OH^-]$

$10^{-10.53}=[OH^-]$

The pH of the solution = ?

$pH+pOH=14$

$pH=14-pOH=14-10.53=3.47$

Here, the pH of the solution is less than 7 which means that solution acidic.

8 0

3 years ago

List the following aqueous solutions in order of decreasing freezing point: 0.040 m glycerin (C3H8O3), 0.020 m potassium bromide

gogolik [260]

Answer:

Solutions from highest to lowest freezing point:

0.040 m glycerin = 0.020 m potassium bromide > 0.030 m phenol

Explanation:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f$ = Depression in freezing point

i = van'T Hoff fcator

$K_f$ = Molal depression constant of solvent

m = molality of the solution

Higher the value of depression in freezing point at lower will be freezing temperature the solution.

1. 0.040 m glycerin

Molal depression constant of water = $K_f=1.86^oC/m$

i = 1 ( organic molecule)

m = 0.040 m

$\Delta T_{f,1}=1\times\times 1.86^oC/m\times 0.040 m$

$\Delta T_{f,1}=0.0744^oC$

2. 0.020 m potassium bromide

Molal depression constant of water = $K_f=1.86^oC/m$

i = 2 (ionic)

m = 0.020 m

$\Delta T_{f,2}=2\times\times 1.86^oC/m\times 0.020 m$

$\Delta T_{f,2}=0.0744^oC$

3. 0.030 m phenol

Molal depression constant of water = $K_f=1.86^oC/m$

i = 1 (organic)

m = 0.030 m

$\Delta T_{f,3}=1\times\times 1.86^oC/m\times 0.030 m$

$\Delta T_{f,3}=0.0558^oC$

$0.0744^oC=0.0744^oC > 0.0558^oC$

$\Delta T_{f,1}=\Delta T_{f,2}>\Delta T_{f,3}$

Solutions from highest to lowest freezing point:

0.040 m glycerin = 0.020 m potassium bromide > 0.030 m phenol

5 0

3 years ago

Which of the following can work against the down-slope pull of gravity?

Phoenix [80]

<span>A. Wind erosion

wind can blow against the force of gravity</span>

8 0

3 years ago

Will ag2so4 precipitate when 100 ml of 5.0×10−2 m agno3 is mixed with 10 ml of 5.0×10−2mna2so4 solution?

Rina8888 [55]

No, it won't. A substance will only precipitate if it is not soluble in an aqueous solution. If you look at the solubility rules, it indicates that all sulfates are soluble except for ions of Ba, Pb, Ca and Sr. Since Ag is not in the list, AgSO4 is soluble and will not precipitate.

4 0

3 years ago

Please help me it’s just that I can’t remember how to do this it’s about the Mole and Stoichiomerty

joja [24]

Answer:

6.52moles of H2O

Explanation:

First, let us calculate the number of moles of O2 containing 1.96x10^24 molecules. This is illustrated below:

1mole of O2 contains 6.02x10^23 molecules.

Therefore Xmol of O2 contain 1.96x10^24 molecules i.e

Xmol of O2 = 1.96x10^24/6.02x10^23 = 3.26moles

Now let us generate a balanced equation for the formation of water. This is illustrated below:

2H2 + O2 —> 2H2O

From the equation,

1mole of O2 produced 2moles of H2O.

Therefore, 3.26moles of O2 will produce = 3.26 x 2 = 6.52moles of H2O

8 0

3 years ago

How do i Round the value 44.981 g to three significant figures.