In a titration, 10.0 ml of 0.0750 M HCl(aq) is exactly neutralized by 30.0 ml of KOH(aq) of unknown concentration. What is the c

Answer:- 0.134 seconds

Solution:- The speed is given as and the circumference is 24900 miles which is same as the distance light have to covered. It asks to calculate the time required to cover this distance by the light.

Unit conversion is needed from miles to meters since the speed is given in meters per second.

1 mile = 1609.34 meters

Thus,

= 40072566 meters

Now,

Rearranged for time, that gives:

Inserting the values:

= 0.134 seconds

Hence, light would take 0.134 seconds to traverse the indicated distance. The answer without the unit is 0.134.

Answer:

710.33 g NO2

Explanation:

2 C8H18 + 25 O2 → 16 CO2 + 18 H2O  

(800 g octane) / (114.2293 g C8H18/mol x (25/2)) = 87.54 mol O2 utilized for combusting octane

= 15.44 mol O2 used for generating NO2

O2 + 2NO → 2NO2

(15.44 mol O2) x (2/2) x (46.0056 g NO2/mol) = 710.33 g NO2

Answer:

Amount of salt in 1 L seawater = 34 g

Explanation:

Based on Archimedes' principle, the mass of fresh water and the mass of the cup are equal to the mass of the same volume of seawater.

The mass of freshwater can be calculated using density times volume.

1 cm³ is equivalent to 1 mL.

The mass of freshwater is 0.999 g/cm³ multiplied by 735 cm³, which results in 734.265 g.

The total mass of the freshwater and cup combined is 734.265 g plus 25 g, equating to 759.265 g.

This means the mass for an equal volume of seawater is 759.265 g.

The volume of the seawater displaced is 735 mL, which is 0.735 L (assuming the cup's volume can be disregarded).

We know that 1 liter equals 1000 cm³ or 1000 mL.

The density of seawater can be determined as mass divided by volume.

The density of seawater becomes 759.265 g divided by 0.735 L, yielding 1033.01 g/L.

Conversely, the density of freshwater in g/L is calculated as 0.999 g/(1/1000) L, equating to 999 g/L.

The mass of salt dissolved in 1 liter of seawater is calculated as 1033.01 g - 999 g, which equals 34.01 g.

Thus, the amount of salt in 1 L of seawater is 34 g.

Answer:

Explanation:

0.3(125)+0.7(126)=37.5+88.2=125.7

2. The most prevalent isotope is 32, since the average is very close to this value.

Response:

K = 6.5 × 10⁻⁶

Detailed Explanation:

C₅H₆O₃ ⇄ C₂H₆ + 3CO

Apply PV=nRT to determine the initial pressure of C₅H₆O₃

P (2.50) = (0.0493) (0.08206) (473)

P = 0.78atm

C₅H₆O₃ ⇄ C₂H₆ + 3CO

0.78atm      0           0

0.78 - x        x           3x

1.63atm = 0.78 - x + x + 3x

P(total) = 0.288atm

C₅H₆O₃ = 0.78 - 0.288

             = 0.489atm

C₂H₆ = 0.288atm

CO = 0.846atm

     = 0.379

   = 6.5 × 10⁻⁶