The Second Law of Thermodynamics makes it difficult for scientists to explain how the original universe became ordered.

Determine the molarity of a 6.0 mole% sulfuric acid solution with SG-a 1.07 Note: Atomic Weight: S (32), O 16); H (O)

marin [14]

Answer:

The molarity of a 6.0 mole% sulfuric acid solution is 2.8157 Molar.

Explanation:

Suppose there are 100 moles in solution:

Moles of sulfuric acid = 6% of 100 moles = 6 moles

Mass of 6 moles of sulfuric acid = 6 mol × 98 g/mol=588 g

Moles of water = 100%- 6% = 94%= 94 moles

Mass of water = 94 mol × 18 g/mol = 1692 g

Specific gravity of the solution ,S.G= 1.07

Density of solution = D

$S.G=\frac{D}{d_w}$

$d_w$ = density of water = 1 g/mL

$D=S.G\times d_w=1.07\times 1 g/mL=1.07 g/mL$

Mass of the solution = 588 g + 1692 g = 2280 g

Volume of the solution = V

Volume = $\frac{Mass}{Density}$

$=\frac{2280 g}{1.07 g/mL}=2130.84 mL=2.13084 L$

1 mL = 0.001 L

$Molarity = \frac{n}{V(L)}$

n = number of moles of compound

V = volume of the solution in L

here we have ,n = 6 moles of sulfuric acid

V = 2.13084 L

So, the molarity of the solution is :

$Molarity=\frac{6 mol}{2.13084 L}=2.8157 mol/L$

5 0

2 years ago

Calculate the solubility of o2 in water at a partial pressure of o2 of 120 torr at 25 ̊c. the henry's law constant for o2 at 25

Vladimir79 [104]

Answer:

1) 2.054 x 10⁻⁴ mol/L.

2) Decreasing the temperature will increase the solubilty of O₂ gas in water.

Explanation:

1) The solubility of O₂ gas in water:

We cam calculate the solubility of O₂ in water using Henry's law: <em>Cgas = K P</em>,

where, Cgas is the solubility if gas,

K is henry's law constant (K for O₂ at 25 ̊C is 1.3 x 10⁻³ mol/l atm),

P is the partial pressure of O₂ (P = 120 torr / 760 = 0.158 atm).

Cgas = K P = (1.3 x 10⁻³ mol/l atm) (0.158 atm) = 2.054 x 10⁻⁴ mol/L.

2) The effect of decreasing temperature on the solubility O₂ gas in water:

Decreasing the temperature will increase the solubilty of O₂ gas in water.

When the temperature increases, the solubility of O₂ gas in water will decrease because the increase in T will increase the kinetic energy of gas particles and increase its motion that will break intermolecular bonds and escape from solution.

Decreasing the temperature will increase the solubility of O₂ gas in water will because the kinetic energy of gas particles will decrease and limit its motion that can not break the intermolecular bonds and increase the solubility of O₂ gas.

6 0

2 years ago

Read 2 more answers

How many grams are in 3.14 x 1015 molecules of CO?

Vesnalui [34]

Answer:

Explanation:

Not Many

1 mol of CO has a mass of

C = 12

O = 16

1 mol = 28 grams.

1 mol of molecules = 6.02 * 10^23

x mol of molecules = 3.14 * 10^15 Cross multiply

6.02*10^23 x = 1 * 3.14 * 10^15 Divide by 6.02*10^23

x = 3.14*10^15 / 6.02*10^23

x = 0.000000005 mols

x = 5*10^-9

1 mol of CO has a mass of 28

5*10^-9 mol of CO has a mass of x Cross Multiply

x = 5 * 10^-9 * 28

x = 1.46 * 10^-7 grams

Answer: there are 1.46 * 10-7 grams of CO if only 3.14 * 10^15 molecules are in the sample

5 0

2 years ago

Susan uses Doppler radar to analyze wind patterns in five cities and determine how windy it is. Which statement describes how sh

andrezito [222]

Answer

Susan can process wind speed data from different regions.

Explanation

A Doppler radar is used in weather forecasting in measuring the direction and speed of objects such as drops of precipitation. It determines if the movement occurring in the atmosphere is horizontally towards or way from the radar. Susan can obtain velocity data about objects at a distance which might be water droplets thus be able to predict a coming weather

5 0

3 years ago

Match the solutions to the descriptions of the freezing points.a. One mole of the ionic compound Na3PO4 dissolved in 1000 g H2O

rosijanka [135]

Explanation:

Depression in Freezing point

= Kf × i × m

where m is molality , i is Van't Hoff factor, m = molality

Since molality and Kf remain the same

depression in freezing point is proportional to i

i= 2 for CuSO4 ( CuSO4----------> Cu+2 + SO4-2

i=1 for C2h6O

i= 3 for MgCl2 ( MgCl2--------> Mg+2+ 2Cl-)

So the freezing point depression is highest for MgCl2 and lowest for C2H6O

so freezing point of the solution = freezing point of pure solvent- freezing point depression

since MgCl2 has got highest freezing point depression it will have loweest freezing point and C2H6O will have highest freezing point

5 0

2 years ago