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wlad13 [49]
3 years ago
14

What is the mass of 5.35 mol of magnesium, Mg?

Chemistry
1 answer:
Hatshy [7]3 years ago
3 0
130.03175 g. Mol to mass is multiplication. So multiply 5.35 by 24.305 and you will get the answer
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C12H26O + SO3+NaOH ----> C12H25NaSO4+ H2O
Vlad [161]

Answer:

We can solve this by the method of which i solved your one question earlier

so again here molar mass of C12H25NaSO4 is 288.372 and number of moles for 11900 gm of C12H25NaSO4 will be = 11900/288.372

which is almost = 41.26 moles

so to get one mole of C12H25NaSO4 we need one mole of C12H26O

so for 41.26 moles of C12H25NaSO4 it will require 41 26 moles of C12H26O

so the mass of C12H26O = 41.26× its molar mass

C12H26O = 41.26×186.34

= 7688.38 gm!!

so the conclusion is If you need 11900 g of C12H25NaSO4 (Sodium Lauryl Sulfate) you need C12H26O 7688.38 gm !!

Again i d k wether it's right or wrong but i tried my best hope it helped you!!

4 0
3 years ago
Explain how the bonding model for sodium metal would differ from the bonding model for sodium chloride, NaCi
Elina [12.6K]
Sodium metal is quite reactive; sodium ions (as in NaCl) are quite unreactive. Cl^- ions are not reactive; they are stingily attracted to positive ions such as Na^+with which they form ionic bonds.
8 0
3 years ago
What is The Magnus Effect/Force? Explain.
cluponka [151]

Answer:

dk

Explanation:

6 0
3 years ago
What differentiates one element from another
worty [1.4K]
I think the number of protons in the nucleus the number of valence electrons atomic mass... 
8 0
3 years ago
Ethylenediamine (en) forms an octahedral complex with Ni2+(aq) with the formula [Ni(en)3]2+. Ni2+(aq) + 3 en ⇌ [Ni(en)3]2+(aq) K
yawa3891 [41]

Answer:

3.125\times 10^{-19} mol/L is the concentration of Ni^{2+}(aq) in the solution.

Explanation:

Ni^{2+}(aq) + 3 en\rightleftharpoons [Ni(en)_3]^{2+}(aq)

Concentration of nickel ion = [Ni^{2+}]=x

Concentration of nickel complex= [[Ni(en)_3]^{2+}]=\frac{0.16 mol}{2 L}=0.08 mol/L

Concentration of ethylenediamine = [en]=\frac{0.80 mol}{2 L}=0.40 mol/L

The formation constant of the complex = K_f=4.0\times 10^{18}

The expression of formation constant is given as:

K_f=\frac{[[Ni(en)_3]^{2+}]}{[Ni^{2+}][en]^3}

4.0\times 10^{18}=\frac{0.08 mol/L}{x\times (0.40 mol/L)^3}

x=\frac{0.08 mol/L}{4.0\times 10^{18}\times (0.40 mol/L)^3}

x=3.125\times 10^{-19} mol/L

3.125\times 10^{-19} mol/L is the concentration of Ni^{2+}(aq) in the solution.

6 0
4 years ago
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