All categories

3 years ago

What is the oh of a solution with a ph of 9.8

Chemistry

1 answer:

Veseljchak [2.6K]3 years ago

4 0

It would be 4.2, hope this helps.

You might be interested in

Please answer asap!

hram777 [196]

1.) A compound is a thing that is composed of two or more (separate) elements.

2.) A compound cannot be separated easily, while its properties differ from where it originates from. It also is formed from chemical reactions.

3.) A pure substance, also known as chemical substance, is a material with constant composition and consistant properties.

4.) If a substance is not a pure substance, then it is an impure substance.

Hope this helped. :)

4 0

3 years ago

A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

Romashka [77]

<u>Answer:</u> The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

<u>Calculation for freezing point of solution:</u>

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

<u>Calculation for boiling point of solution:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

8 0

3 years ago

13.50 grams of Pb(NO3)4 are dissolved in enough water to make 250 mL of solution. What is the molar its of the resulting solutio

likoan [24]

Data:
M (molarity) = ? (M or Mol/L)
m (mass) = 13.50 g
V (volume) = 250 mL → 0.25 L
MM (Molar Mass) of Lead(IV) Nitrate $Pb(NO_3)_4$
Pb = 1*207 = 207 amu
N = (1*14)*4 = 14*4 = 56 amu
O = (3*16)*4 = 48*4 = 192 amu
------------------------------------
MM of $Pb(NO_3)_4$ = 207+56+192 = 455 g/mol

Formula:
$M = \frac{m}{MM*V}$

Solving:
$M = \frac{m}{MM*V}$
$M = \frac{13.50}{455*0.25}$
$M = \frac{13.50}{113.75}$
$M = 0.118681318...\:\:\to\:\:\boxed{\boxed{M \approx 0.119\:Mol/L}}\end{array}}\qquad\quad\checkmark$

Answer:
<span>B. 0.119 M</span>

5 0

3 years ago

What mass of magnesium chloride is needed to make 100.0 mL of a solution that is 0.500 M in chloride ion?

miss Akunina [59]

M = n/V

.5M = n/.100 L

n = .1 L * .5M

n= .05 mols of MgCl2

mass of MgCl2 = .05 mols of MgCl2 * 95.211 grams/ 1 mol of MgCl2

mass of MgCl2 = 4.76 grams

4.76 grams of MgCl2 is needed to make 100 ml of a solution that is .500M, in chloride ion. Bolded = confused

3 0

3 years ago

What is the mass of 4.99×1021 platinum atoms?

charle [14.2K]

Answer:

$\boxed {\boxed {\sf 1.62 \ g \ Pt}}$

Explanation:

We are asked to find the mass of a number of platinum (Pt) atoms.

<h3>1. Convert Atoms to Moles </h3>

First, we convert atoms to moles using Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are platinum atoms.

We will convert using dimensional analysis so we set up a ratio using the information we know (6.022 × 10²³ platinum atoms in 1 mole of platinum).

$\frac {6.022 \times 10^{23} \ atoms \ Pt}{ 1 \ mol \ Pt}$

We are converting 4.99 ×10²¹ atoms of Pt to moles of Pt, so we multiply by this value.

$4.99 \times 10^{21} \ atoms \ Pt *\frac {6.022 \times 10^{23} \ atoms \ Pt}{ 1 \ mol \ Pt}$

Flip the ratio so the units of atoms of platinum cancel.

$4.99 \times 10^{21} \ atoms \ Pt *\frac { 1 \ mol \ Pt}{6.022 \times 10^{23} \ atoms \ Pt}$

$4.99 \times 10^{21} *\frac { 1 \ mol \ Pt}{6.022 \times 10^{23}}$

$\frac { 4.99 \times 10^{21}}{6.022 \times 10^{23}} \ mol \ Pt$

Divide.

$0.008286283627 \ mol \ Pt$

<h3>2. Convert Moles to Grams </h3>

Next, we convert moles to grams using the molar mass. This is the mass of 1 mole of a substance. This is found on the Periodic Table because it is equivalent to the atomic mass, but the units are grams per mole instead of atomic mass units. Look up platinum's molar mass.

Pt: 195.08 g/mol

Set up another ratio using this new information (195.08 grams of Pt in 1 mole of Pt).

$\frac {195.08 \ g \ Pt}{ 1 \ mol \ Pt}$

Multiply by the number of moles we just calculated.

$0.008286283627 \ mol \ Pt*\frac {195.08 \ g \ Pt}{ 1 \ mol \ Pt}$

The units of moles of platinum cancel.

$0.008286283627*\frac {195.08 \ g \ Pt}{ 1 }$

$0.008286283627* {195.08 \ g \ Pt}$

$1.61648821\ g \ Pt$

<h3>3. Round</h3>

The original measurement of atoms ( 4.99 ×10²¹ ) has 3 significant figures, so our answer must have the same. For the number we found, that is the hundredth place. The 6 in the thousandth place tells us to round the 1 up to a 2.

$1.62 \ g \ Pt$

There are approximately <u>1.62 grams of platinum</u> in 4.99 ×10²¹ atoms of platinum.

8 0

3 years ago