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IceJOKER [234]
2 years ago
7

2. A compound with the following composition by mass: 24.0% C, 7.0% H, 38.0% F, and 31.0% P. what is the empirical formula

Chemistry
1 answer:
Svetach [21]2 years ago
5 0

Answer:

C₂H₇F₂P

Explanation:

Given parameters:

Composition by mass:

                C = 24%

                H = 7%

                 F  = 38%

                 P  = 31%

Unknown:

Empirical formula of compound;

Solution :

The empirical formula is the simplest formula of a compound. To solve for this, follow the process below;

                                   C                          H                         F                   P

% composition

by mass                     24                          7                        38                  31

Molar mass                 12                           1                         19                  31

Number of

moles                       24/12                          7/1                    38/19           31/31

                                     2                               7                       2                   1

Dividing

by the

smallest                      2/1                             7/1                       2/1                1/1

                                     2                                7                        2                   1

           Empirical formula        C₂H₇F₂P

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Answer: 178.9 g

Explanation:

Density = \frac{mass}{volume in mL}

find volume of the cube: (5.80 cm) (5.80 cm) (5.80cm) = 195.112 cm³

1.0 cm³ = 1.0 mL

so 195.112 cm³ = 195.112 mL

plug value into density equation:

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3 0
1 year ago
What is the number of moles in 15.0 g AsH3?
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Explanation:

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<em>n = mass / molar mass.</em>

mass of AsH₃ = 15.0 g.

molar mass of AsH₃ = 77.95 g/mol.

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7 0
1 year ago
DNA can be replicated through an in vivo mechanism or a polymerase chain reaction (PCR) mechanism. Match each description to the
umka2103 [35]

Complete question from other source attached

Answer:

Explanation:

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7 0
2 years ago
If 4.168 kJ of heat is added to a calorimeter containing 75.40 g of water, the temperature of the water and the calorimeter incr
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Answer:

The value of  the heat capacity of the Calorimeter  C_c = 54.4 \frac{J}{c}

Explanation:

Given data

Heat added Q = 4.168 KJ = 4168 J

Mass of water m_w = 75.40 gm

Temperature change = ΔT = 35.82 - 24.58 = 11.24 ° c

From the given condition

Q = m_w C_w ΔT + C_c ΔT

Put all the values in above equation we get

4168 = 75.70 × 4.18 × 11.24 +  C_c × 11.24

611.37 =  C_c × 11.24

C_c = 54.4 \frac{J}{c}

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7 0
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Since these must equal, 0.57213 = 297.5/T
T = 519.98 K
Subtracting 273.15 gives 246.83 degC.
5 0
1 year ago
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