Nonpolar and small polar molecules can pass through the cell membrane, so they diffuse across it in response to concentration gradients. Carbon dioxide and oxygen are two molecules that undergo this simple diffusion through the membrane. The simple diffusion of water is known as osmosis.
984 grams of strontium will be recovered from 9.84x10^8 cubic meter of seawater.
Explanation:
From the question data given is :
volume of strontium in sea water= 9.84x10^8 cubic meter
(1 cubic metre = 1000000 ml)
so 9 .84x10^8 cubic meter
= 984 ml.
density of sea water = 1 gram/ml
from the formula mass of strontium can be calculated.
density = 
mass = density x volume
mass = 1 x 984
= 984 grams of strontium will be recovered.
98400 centigram of strontium will be recovered.
Strontium is an alkaline earth metal and is highly reactive.
<h3>
Answer:</h3>
0.111 J/g°C
<h3>
Explanation:</h3>
We are given;
- Mass of the unknown metal sample as 58.932 g
- Initial temperature of the metal sample as 101°C
- Final temperature of metal is 23.68 °C
- Volume of pure water = 45.2 mL
But, density of pure water = 1 g/mL
- Therefore; mass of pure water is 45.2 g
- Initial temperature of water = 21°C
- Final temperature of water is 23.68 °C
- Specific heat capacity of water = 4.184 J/g°C
We are required to determine the specific heat of the metal;
<h3>Step 1: Calculate the amount of heat gained by pure water</h3>
Q = m × c × ΔT
For water, ΔT = 23.68 °C - 21° C
= 2.68 °C
Thus;
Q = 45.2 g × 4.184 J/g°C × 2.68°C
= 506.833 Joules
<h3>Step 2: Heat released by the unknown metal sample</h3>
We know that, Q = m × c × ΔT
For the unknown metal, ΔT = 101° C - 23.68 °C
= 77.32°C
Assuming the specific heat capacity of the unknown metal is c
Then;
Q = 58.932 g × c × 77.32°C
= 4556.62c Joules
<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
- We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
4556.62c Joules = 506.833 Joules
c = 506.833 ÷4556.62
= 0.111 J/g°C
Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C
Answer:
you would be trapped in the gravitational pull of a black hole
Explanation:
there is a black hole in the center of the galacy
