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rosijanka [135]
2 years ago
12

Given that the molar mass of H2O is 18.02 g/mol, how many liters of propane are required at STP to produce 75 g of H2O from this

reaction?
23 L23.3 L93 L93.2 L
Chemistry
1 answer:
alukav5142 [94]2 years ago
4 0
First, we require the combustion equation of propane:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

We see that each mole of propane produces 4 moles of water.
Now, we calculate the moles of water required: 75 / 18.02
Moles required = 4.16
Moles of propane required for this are: 4.16 / 4 = 1.04 moles

Each mole of gas occupies 22.14 liters, so the volume of propane required are 1.04 * 22.14 = 23.03 L
23 L of propane are required
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The crystalline hydrate cd(no3)2 ⋅ 4h2o(s) loses water when placed in a large, closed, dry vessel at room temperature: cd(no3)2⋅
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The given dehydration equation is,

Cd(NO_{3})_{2}. 4H_{2}O (s) ---> Cd(NO_{3})_{2}(s) + 4 H_{2}O(g)

Cadmiumnitrate tetrahydrate when heated dehydrates releasing the combined water as water vapor. The reaction produces 4 moles of gaseous product water vapor. So, the degree of disorder or randomness increases. Hence, the sign of change in entropy is positive.

This reaction is spontaneous at room temperature even if it is endothermic as the sign of change in entropy is positive.

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2 years ago
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What is the element with the nlx notation of 5d2?
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7 0
2 years ago
Determine the specific heat (in J/g C) for a 2.508 kilogram substance which increases its temperature from 4.051 C to 42.061 C w
just olya [345]

Answer: 0.036 J/g°C

Explanation:

The quantity of heat energy (Q) required to raise the temperature of a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Given that,

Q = 3.42 Kilojoules

[Convert 3.42 kilojoules to joules

If 1 kilojoule = 1000 joules

3.42 kilojoules = 3.42 x 1000 = 3420J]

Mass = 2.508Kg

[Convert 2.508 kg to grams

If 1 kg = 1000 grams

2.508kg = 2.508 x 1000 = 2508g]

C = ? (let unknown value be Z)

Φ = (Final temperature - Initial temperature)

= 42.061°C - 4.051°C

= 38.01°C

Apply the formula, Q = MCΦ

3420J = 2508g x Z x 38.01°C

3420J = 95329.08g•°C x Z

Z = (3420J / 95329.08g•°C)

Z = 0.03588 J/g°C

Round the value of Z to the nearest thousandth, hence Z = 0.036 J/g°C

Thus, the specific heat of the substance is 0.036 J/g°C

7 0
2 years ago
Draw a sodium formate molecule. The structure has been supplied here for you to copy. To add formal charges, click the button be
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The Molecule of Sodium Formate along with Formal Charges (in blue) and lone pair electrons (in red) is attached below.

Sodium Formate is an ionic compound made up of a positive part (Sodium Ion) and a polyatomic anion (Formate).

Nomenclature:

                       In ionic compounds the positive part is named first. As sodium ion is the positive part hence, it is named first followed by the negative part i.e. formate.

Name of Formate:

                             Formate ion has been derived from formic acid ( the simplest carboxylic acid). When carboxylic acids looses the acidic proton of -COOH, they are converted into Carboxylate ions.

E.g.

                    HCOOH (formic acid)    →     HCOO⁻ (formate)  +  H⁺

                H₃CCOOH (acetic acid)     →      H₃CCOO⁻ (acetate)  +  H⁺

Formal Charges:

                           Formal charges are calculated using following formula,

          F.C  =  [# of Valence e⁻] - [e⁻ in lone pairs + 1/2 # of bonding electrons]

For Oxygen:

                    F.C  =  [6] - [6 + 2/2]

                    F.C  =  [6] - [6 + 1]

                    F.C  =  6 - 7

                    F.C  =  -1

For Sodium:

                    F.C  =  [1] - [0 + 0/2]

                    F.C  =  [1] - [0]

                    F.C  =  1 - 0

                    F.C  =  +1

5 0
2 years ago
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