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uysha [10]
2 years ago
5

Find the percent composition for SF6 S =? % F =? %

Chemistry
1 answer:
goldenfox [79]2 years ago
6 0

Answer:

S = 21.92 %

F = 78.08 %

Explanation:

To find the percent composition of each element in SF6, we must find the molar mass of SF6 first.

Molar mass of SF6 = 32 + 19(6)

= 32 + 114

= 146g/mol

mass of Sulphur (S) in SF6 = 32g

mass of Fluorine (F) in SF6 = 114g

Percent composition = mass of element/molar mass of compound × 100

- % composition of S = 32/146 × 100 = 21.92%.

- % composition of F = 114/146 × 100 = 78.08%.

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You need 1.2 moles of H2SO4 for an experiment. You weigh out 100g of this colorless syrupy liquid. Do you have enough?
Sladkaya [172]
Answer is: not enough <span>colorless syrupy liquid.
</span>n(H₂SO₄) = 1,2 mol.
M(H₂SO₄) = 2Ar(H) + Ar(S) + 4Ar(O) · g/mol.
M(H₂SO₄) = 2·1 + 32 + 4·16 · g/mol.
M(H₂SO₄) = 98 g/mol.
m(H₂SO₄) = n(H₂SO₄) · M(H₂SO₄).
m(H₂SO₄) = 1,2 mol · 98 g/mol.
m(H₂SO₄) = 117,6 g needed.
100 g is less that 117,6 g.
8 0
3 years ago
strontium-90 has a half life of 29 years. if a site held 4000 kg of this isotope, approximately what mass of strontium-90 would
Ludmilka [50]

Answer:

125g

Explanation:

4 0
2 years ago
A radioactive nucleus emits a beta particle, then the parent and daughter nuclei are
Nana76 [90]

Answer:

isobars

Explanation:

How?

  • A radioactive nucleus emits beta particle(Like uranium,radium)
  • So the mass numbers are same for daughter nuclei .
  • They have different atomic numbes .

So they are isobars

3 0
2 years ago
Read 2 more answers
Identify the solute with the highest van't Hoff factor. And how do you determine which one is highest?
german

Answer : The correct option is, (A) AlCl_3

Explanation :

Van't Hoff factor : It is defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property.

We can determine the van't Hoff factor by the association and dissociation of the compound.

(A) AlCl_3

It is an electrolyte that dissociates to give aluminum ion and chloride ion.

The dissociation of AlCl_3 will be,

AlCl_3\rightarrow Al^{3+}+3Cl^{-}

So, Van't Hoff factor = Number of solute particles = Al^{3+}+3Cl^{-} = 1 + 3 = 4

(B) KI

It is an electrolyte that dissociates to give potassium ion and iodide ion.

The dissociation of KI will be,

KI\rightarrow K^{+}+I^{-}

So, Van't Hoff factor = Number of solute particles = K^{+}+I^{-} = 1 + 1 = 2

(C) CaCl_2

It is an electrolyte that dissociates to give calcium ion and chloride ion.

The dissociation of CaCl_2 will be,

CaCl_2\rightarrow Ca^{2+}+2Cl^{-}

So, Van't Hoff factor = Number of solute particles = Ca^{2+}+2Cl^{-} = 1 + 2 = 3

(D) MgSO_4

It is an electrolyte that dissociates to give magnesium ion and sulfate ion.

The dissociation of MgSO_4 will be,

MgSO_4\rightarrow Mg^{2+}+SO_4^{2-}

So, Van't Hoff factor = Number of solute particles = Mg^{2+}+SO_4^{2-} = 1 + 1 = 2

(E) Non-electrolyte

The dissociation non-electrolyte is not possible. So, the Van't Hoff factor will always be, 1.

Hence, the highest van't Hoff factor of solute is, AlCl_3

7 0
3 years ago
Oxygen Supply in Submarines Nuclear submarines can stay under water nearly indefinitely because they can produce their own oxyge
andrew11 [14]

Answer:

0.025 L

Explanation:

The production of oxygen in the electrolysis of water is;

4OH^-(aq) -----> 2H2O(l) + O2(g) + 4e

Since 1 F = 96500C

molar volume of a gas = 22.4 L

From the reaction equation;

4 * 96500 C yields 22.4 L of oxygen

(3 * 60 * 60 *  0.0400) C yields (3 * 60 * 60 *  0.0400) * 22.4/4 * 96500

= 9676.8/386000

= 0.025 L

3 0
3 years ago
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