Ask question

Ask question

All categories

AleksandrR [38]

4 years ago

11

When an object such as a rock is dropped into water it disturbs the surface of the water waves form at the surface of the water

and travel outward from the point of the disturberance

1 answer:

viktelen [127]4 years ago

3 0

So this isn't a question but thanks for telling me...? Next time ask a question pleaseeeee.

You might be interested in

Calculating Density Warm Up

Artist 52 [7]

Answer:

2

Explanation:

33-25=8

48/8=6

6 0

4 years ago

Astatine, At, is below iodine in Group 17 of the Periodic Table. Which statement is most likely to be correct?

Sphinxa [80]

The answer is A hope this helps

8 0

3 years ago

What effect does rapid exhalation of CO2 during exercise have on the concentration of H2CO3 in the blood?

BARSIC [14]

If there is a constant loss of $CO_{2}~(carbon~dioxide)$ the concentration of $H_{2} CO_{3}~(carbonic~acid)$ will be affected and decrease since $CO_{2}$ is a main component of $H_{2}CO_{3}$ .

Hope it helped,

BioTeacher101

3 0

3 years ago

Read 2 more answers

Evaporate minerals, or "salts," include halite, anhydrite, and ______.

pav-90 [236]

B is the correct answer! I learned this in class last week :)

5 0

3 years ago

Read 2 more answers

Hexane and octane are mixed to form a 45 mol% hexane solution at 25 deg C. The densities of hexane and octane are 0.655 g/cm3 an

avanturin [10]

Answer:

The required volume of hexane is 0.66245 Liters.

Explanation:

Volume of octane = v=1.0 L= $1000 cm^3$

Density of octane= d = $0.703 g/cm^3$

Mass of octane ,m= $d\times v=0.703 g/cm^3\times 1000 cm^3=703 g$

Moles of octane = $\frac{m}{114 g/mol}=\frac{703 g}{114 g/mol}=6.166 mol$

Mole percentage of Hexane = 45%

Mole percentage of octane = 100% - 45% = 55%

$55\%=\frac{6.166 mol}{\text{Total moles}}\times 100$

Total moles = 11.212 mol

Moles of hexane :

$45%=\frac{\text{moles of hexane }}{\text{Total moles}}\times 100$

Moles of hexane = 5.0454 mol

Mass of 5.0454 moles of hexane,M = 5.0454 mol × 86 g/mol=433.9044 g

Density of the hexane,D = $0.655 g/cm^3$

Volume of hexane = V

$V=\frac{M}{D}=\frac{433.9044 g}{0.655 g/cm^3}=662.4494 cm^3\approx 0.66245 L$

(1 cm^3= 0.001 L)

The required volume of hexane is 0.66245 Liters.

5 0

3 years ago