Answer:
0.15 L
Explanation:
You need to first find the volume of the container. You can do this by dividing the mass by the density. This will give you the mass in mL.
5.00 kg = 5,000 g
(5,000 g)/(1.00 g/mL) = 5,000 mL
5,000 mL = 5 L
Now, find the volume the seawater will take up.
(5,000 g)(1.03 g/mL) = 4854.4 mL
4854.4 mL = 4.85 L
Subtract the two volumes to find the volume that left unfilled.
5 L - 4.85 L = 0.15 L
Density = mass/volume
so rearranged mass = volume x density
mass = 8.920 x 45 = 401.4g
rearrange (there are 1000grams in 1kg)
volume = mass/density
volume = 1000/8.920
volume = 112.1076233cm3
Answer:
The correct answer is A) circulatory
Explanation:
The circulatory system collects metabolic wastes from the blood, which will be eliminated by the kidneys through urine.
The kidneys are responsible for the elimination of waste, acids and excess fluid from the body, maintaining the balance of water, salts and minerals. Blood flows into the kidney through the renal artery.
Answer:
63. 55 amu
Explanation:
Copper is known to exist in two different isotopes which are Cu-63 and Cu-65.
Cu-63 has an atomic mass of 62.93 amu and it has an abundance of 69.15%.
Similarly,
Cu-65 has an atomic mass of 64.93 amu and it has an abundance of 30.85%
Therefore, using the weighted average mass method, the atomic mass of copper is:
Atomic mass of copper = (0.6915*62.93) amu + (0.3085*64.93) amu = 43.52 amu + 20.03 amu = 63.55 amu
Thus, the atomic mass of copper (express in two decimal places) is 63.55 amu
Answer:
Na₁₁ = 1s² 2s² 2p⁶ 3s¹
Explanation:
Sodium is present in group 1.
It is alkali metal.
It has one valence electron.
The atomic number of sodium is 11.
Its atomic mass is 23 amu.
The longhand notation of electronic configuration of sodium can be written as,
Na₁₁ = 1s² 2s² 2p⁶ 3s¹
The electronic configuration in shorthand notation( noble gas) would be written as,
Na₁₁ = [Ne] 3s¹
Sodium loses its one valence electron to complete the octet and get stable thus form +1 cation.
It react with halogen and form salt. Such as sodium chloride.
2Na + Cl₂ → 2NaCl
