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adoni [48]
3 years ago
7

In which reaction is it possible for nitrogen to change into carbon?

Chemistry
1 answer:
sertanlavr [38]3 years ago
5 0
C. ionization reaction...
THANKS!!
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Zinc, sulphur, oxygen<br> What would this compound be called?
Nataliya [291]

Zinc sulphate

Zinc sulphateZnSO₄

This is the answer.

6 0
2 years ago
What is the difference between atomic mass and molar mass?
sweet-ann [11.9K]
Atomic mass is the mass of a chemical element expressed in atomic mass units. Molar mass the mass of a given substance divided by its amount of substance in (mol) .
6 0
3 years ago
Complex chemical catalysts produced by living cells are called enzymes. true false
SIZIF [17.4K]
True 
Enzymes help the body in digestion and other bodily functions which involve chemicals.
7 0
3 years ago
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Determine the relative formula mass of hexasodium difluoride using the periodic table below. A. 138 g/mol B. 176 g/mol C. 20 g/m
laila [671]

Answer:

Option B. 176g/mol

Explanation:

We'll begin by writing the chemical formula for hexasodium difluoride. This is given below:

Hexasodium means 6 sodium atom

Difluoride means 2 fluorine atom.

Therefore, the formula for hexasodium difluoride is Na6F2.

The relative formula mass of a compound is obtained by simply adding the atomic masses of the elements present in the compound.

Thus, the relative formula mass of hexasodium difluoride, Na6F2 can be obtained as follow:

Molar mass of Na = 23g/mol

Molar mass of F = 19g/mol

Relative formula mass Na6F2 = (23x6) + (19x2)

= 138 + 38

= 176g/mol

Therefore, the relative formula mass of hexasodium difluoride, Na6F2 is 176g/mol

3 0
3 years ago
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You are given a solid that is a mixture of na2so4 and k2so4.
Murljashka [212]

Here we have to calculate the amount of SO_{4}^{2-} ion present in the sample.

In the sample solution 0.122g of SO_{4}^{2-} ion is present.

The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-

K₂SO₄ = 2K⁺ +  SO_{4}^{2-}

(Na)₂SO₄=2Na⁺ +  SO_{4}^{2-}

Thus, BaCl₂+  SO_{4}^{2-} = BaSO₄↓ + 2Cl⁻ .

(Na)₂SO₄ and  K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of  SO_{4}^{2-} ion is precipitated in this reaction.  

The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.

Now the molecular weight of BaSO₄ is 233.3 g/mol.

We know the molecular weight of sulfate ion (SO_{4}^{2-}) is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of SO_{4}^{2-} ion is present.

Or. we may write in 233.3 g of BaSO₄ 96.06 g of SO_{4}^{2-} ion is present. So in 1 g of BaSO₄ \frac{96.06}{233.3}=0.411 g of SO_{4}^{2-} ion is present.

Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of SO_{4}^{2-} ion is present.        

5 0
3 years ago
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