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QveST [7]
3 years ago
7

A sample of a compound contains 3.21 g of sulfur and 11.4 g of fluorine. Which of the following represents the empirical formula

of the compound?
(A) SF2 (B) SF3 (C) SF4 (D) SF5 (E) SF6
Chemistry
1 answer:
Sergeeva-Olga [200]3 years ago
7 0

Answer:

The empirical formula is SF6 (option E)

Explanation:

Step 1: Data given

Mass of sulfur = 3.21 grams

Mass of fluorine = 11.4 grams

Molar mass sulfur = 32.065 g/mol

Molar mass fluorine = 19.00 g/mol

Step 2: Calculate moles

Moles = mass /molar mass

Moles sulfur = 3.21 grams / 32.065 g/mol

Moles sulfur = 0.100 moles

Moles fluorine = 11.4 grams / 19.00 g/mol

Moles fluorine = 0.600 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

S: 0.100 / 0.100 = 1

F : 0.600 / 0.100 = 6

The empirical formula is SF6 (option E)

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If I have 340 mL of a 1.5 M NaBr solution, what will the concentration be if I add 560 mL more water to it?
ipn [44]

Answer:

0.5667 M ≅ 0.57 M.

Explanation:

It is known that the no. of millimoles of a solution before dilution is equal to the no. of millimoles of the solution after the dilution.

It can be expressed as:

(MV) before dilution = (MV) after dilution.

M before dilution = 1.5 M, V before dilution = 340 mL.

M after dilution = ??? M, V after dilution = 340 mL + 560 mL = 900 mL.

∴ M after dilution = (MV) before dilution/(V) after dilution = (1.5 M)(340 mL)/(900 mL) = 0.5667 M ≅ 0.57 M.

5 0
2 years ago
What is the volume in liters of 5.25 moles of He gas?​
serious [3.7K]

Answer:

\boxed {\boxed {\sf 117.6 \ L \ He}}

Explanation:

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We can set up a ratio.

\frac { 22.4 \ L \ He}{ 1 \ mol \ He}

Multiply by the given number of moles.

5.25 \ mol \ He *\frac { 22.4 \ L \ He}{ 1 \ mol \ He}

The moles of helium will cancel.

5.25 *\frac { 22.4 \ L \ He}{ 1 }

5.25 * { 22.4 \ L \ He}

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117.6 \ L \ He

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again  at  STP  1mole =  22.4 L  what  about 1.876 moles

=    22.4 L x 1.876  moles/ 1 mole =  42.02 L


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