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mariarad [96]
3 years ago
7

A 1.11 g sample of caffeine, C8H10N4O2, burns in a constant-volume calorimeter that has a heat capacity of 7.85 kJ/K. The temper

ature increases from 298.70 K to 303.81 K. What is the molar heat of combustion of caffeine (in kJ).?

Chemistry
1 answer:
VladimirAG [237]3 years ago
4 0

Answer:

401135 kJ

Explanation:

From the balanced quation,

(q/n) = CΔE

Molar heat of combustion = 7.85kJk × (303.81-298.70)k

                                            = 7.85kj × 5.11

                                             = 40.1135kj

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How many grams of CO₂ can be produced from the combustion of 2.76 moles of butane according to this equation: 2 C₄H₁₀(g) + 13 O₂
Vilka [71]

Answer:

485.76 g of CO₂ can be made by this combustion

Explanation:

Combustion reaction:

2 C₄H₁₀(g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (g)

If we only have the amount of butane, we assume the oxygen is the excess reagent.

Ratio is 2:8. Let's make a rule of three:

2 moles of butane can produce 8 moles of dioxide

Therefore, 2.76 moles of butane must produce (2.76 . 8)/ 2 = 11.04 moles of CO₂

We convert the moles to mass → 11.04 mol . 44g / 1 mol = 485.76 g

5 0
3 years ago
The air in a 2 L balloon at 0.998 atm and 34.0 °C. What will be its pressure if it is brought to a higher altitude where it now
Phantasy [73]

Answer: 0.529 atm

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 0.998 atm

P_2 = final pressure of gas = ?

V_1 = initial volume of gas = 2 L

V_2 = final volume of gas = 3.5 L

T_1 = initial temperature of gas = 34.0^oC=273+34.0=307.0K

T_2 = final temperature of gas = 12.0^oC=273+12.0=285.0K

Now put all the given values in the above equation, we get:

\frac{0.998\times 2}{307.0K}=\frac{P_2\times 3.5}{285.0K}

P_2=0.529atm

Thus the pressure if it is brought to a higher altitude where it now occupies 3.5 L and is at 12.0 °C is 0.529 atm

4 0
3 years ago
A three-carbon alkene contains two double bonds. Predict its molecular formula? A) C3H4 B) C3H5 C) C3H6
anyanavicka [17]
A three-carbon alkene contains two double bonds. My prediction of its molecular formula is A. C3H4.
5 0
3 years ago
Read 2 more answers
Give the hybridization for the c in hcch. group of answer choices
Tamiku [17]

The hybridization for C in acetylene, HCCH, or C₂H₂ is 'sp'.

Discussion:

There are three different forms of hybridization -

  1. sp- The first occurs when two carbon atoms are triple linked.
  2. sp₂- When two carbon atoms are double-bonded to one another, this is known as sp₂.
  3. sp₃- When a single bond joins two carbon atoms, this is known as sp₃.

In the case of acetylene(HCCH or C₂H₂):

  • The carbon atom requires additional electrons to establish four bonds with hydrogen and other carbon atoms in the synthesis of C₂H₂. One 2s₂ pair is consequently transferred to the vacant 2pz orbital. Each carbon has two sp hybrid orbitals after the 2s orbital in each atom combines with one of the 2p orbitals.
  • As a result of the atoms' symmetrical alignment in a single plane, C₂H₂ possesses a linear molecular structure. Due to their lower electronegative nature than Hydrogen atoms, all Carbon atoms are situated near the center of the Lewis structure of C₂H₂.

                                              H-C≡C-H

Therefore, it is concluded from the above discussion that the hybridization type of acetylene is 'sp'.

Learn more about hybridization here:

brainly.com/question/14140731

#SPJ4

5 0
2 years ago
صف زوايا NO و بالترتيب من الأصغر إلى الأكبر، 9<br>و12 - OM<br>199​
scoundrel [369]
I CANT TRANSLATE IT AND IT WONT LET ME COPY ON MOBILE. I WOULD HELP U IF YOU DIDNT WRITE IT IN ARABIC
8 0
2 years ago
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