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lakkis [162]
3 years ago
7

What is the average salinity of ocean water

Chemistry
2 answers:
pentagon [3]3 years ago
5 0
3.45g/kg 
Hope I helped!
Georgia [21]3 years ago
5 0

its 34.5 i took the test

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What causes the part of earths magnetic field called the magnetosphere to exist?
Sav [38]

Answer:The solar wind creates the magnetosphere as it pushes against and shapes Earth's magnetic field.

Explanation:

4 0
3 years ago
What is the mass, in grams, of a pure gold cube that has a volume of 4.30 cm3?
umka21 [38]
The density of a pure gold is 19.32g/cm3
if the volume is 4.30 cm3 then we can calculate the mass using the formula mass=density x volume
mass=19.32*4.30
mass=83.076g
4 0
3 years ago
Convert 18.9 moles to MgCl2 to formula units
Masja [62]

Answer:

18.9 moles of MgCl2 = 17.834 kg of MgCl2

Explanation:

The molecular weight of MgCl is 80.0 g/mol . So, to convert the given mole amount to grams, multiply this by this number, which is constant for all compounds with a specific composition (mass fraction).

Considering the original question was in the context of chemistry, I wanted to make it seem formal and more educational too. Hopefully that worked!  

EDIT: Came up with some text around what happens inside cells that would have made it better if someone just had an issue converting units, but I doubt my answer will be accepted >.<

3 0
2 years ago
Magnesium acetate can be prepared by a reaction involving 15.0 grams of iron(III) acetate with either 10.0 grams of Magnesium Ch
noname [10]
1) Chemical reaction:
2(CH₃COO)₃Fe + 3MgCrO₄ → Fe₂(CrO₄)₃ + 3(CH₃COO)₂Mg.
m((CH₃COO)₃Fe) = 15,0 g.
m(MgCrO₄) = 10,0 g.
n((CH₃COO)₃Fe) = m((CH₃COO)₃Fe) ÷ M((CH₃COO)₃Fe).
n((CH₃COO)₃Fe) = 15 g ÷ 233 g/mol.
n((CH₃COO)₃Fe) = 0,064 mol.
n(MgCrO₄) = m(MgCrO₄) ÷ M(MgCrO₄).
n(MgCrO₄) = 10 g ÷ 140,3 g/mol.
n(MgCrO₄) = 0,071 mol; limiting reagens.
From chemical reaction: n(MgCrO₄) : n((CH₃COO)₂Mg) = 3 : 3.
n((CH₃COO)₂Mg) = 0,071 mol.
m((CH₃COO)₂Mg) = 0,071 mol · 142,4 g/mol.
n((CH₃COO)₂Mg) = 10,11 g.

2) Chemical reaction: 
2(CH₃COO)₃Fe + 3MgSO₄ → Fe₂(SO₄)₃ + 3(CH₃COO)₂Mg.
m((CH₃COO)₃Fe) = 15,0 g.
m(MgSO₄) = 15,0 g.
n((CH₃COO)₃Fe) = m((CH₃COO)₃Fe) ÷ M((CH₃COO)₃Fe).
n((CH₃COO)₃Fe) = 15 g ÷ 233 g/mol.
n((CH₃COO)₃Fe) = 0,064 mol; limiting ragens.
n(MgSO₄) = m(MgSO₄) ÷ M(MgSO₄).
n(MgSO₄) = 15 g ÷ 120,36 g/mol.
n(MgSO₄) = 0,125 mol; limiting reagens.
From chemical reaction: n(CH₃COO)₃Fe) : n((CH₃COO)₂Mg) = 2 : 3.
n((CH₃COO)₂Mg) = 0,064 mol · 3/2.
n((CH₃COO)₂Mg) = 0,096 mol.
m((CH₃COO)₂Mg) = 0,096 mol · 142,4 g/mol.
m((CH₃COO)₂Mg) = 13,7 g.
7 0
3 years ago
Is oxidation state of metals in amalgams 0 .If yes why ?
Alisiya [41]
Hi Yes the oxidation state of metal in amalgam is zero. I will say it's because the metals are in their elemental state. So I hope that's help !
7 0
3 years ago
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