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77julia77 [94]
3 years ago
5

Bananas Foster is an example of a dessert that is flambéed. A Bananas Foster label states the accepted number of Calories to be

only 300 calories, but a calorimetry experiment measured there to be 318 calories. Calculate the percent error.
Chemistry
1 answer:
WINSTONCH [101]3 years ago
4 0

Answer:

6.00%

Explanation:

Step 1: Given data

Accepted value for the number of calories in a Bananas Foster: 300 calories

Measured value for the number of calories in a Bananas Foster: 318 calories

Step 2: Calculate the percent error in the measure

We will use the following expression.

%error = |accepted value - experimental value|/ accepted value × 100%

%error = |300 cal - 318 cal|/ 300 cal × 100% = 6.00%

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Two solutions, initially at 24.60 °C, are mixed in a coffee cup calorimeter (Ccal = 15.5 J/°C). When a 100.0 mL volume of 0.100
yulyashka [42]

Answer:

ΔH = -59.6kJ/mol

Explanation:

The reaction that occurs between Ag⁺ and Cl⁻ ions is:

Ag⁺ + Cl⁻ → AgCl(s) + ΔH

To find ΔH we need to obtain moles of reaction and heat released in the reaction because ΔH is defined as heat released per mole of reaction.

<em>Moles of reaction:</em>

Moles of Ag⁺ and Cl⁻ added are:

Ag⁺: 0.100L * (0.100mol / L) = 0.01moles

Cl⁻: 0.100L * (0.200mol / L) 0 0.02 moles

That means limiting reactant is Ag⁺ and moles of reaction are 0.01 moles

<em>Heat released:</em>

To find heat released we must use coffe cup calorimeter equation:

Q = C*m*ΔT

<em>Where C is specific heat of solution (4.18J/g°C), m is the mass of solution (200g because there are 100 + 100mL = 200mL and density of solution is 1g/mL) and ΔT is change in temperature (25.30°C - 24.60°C = 0.70°C).</em>

Replacing:

Q = C*m*ΔT

Q = 4.18J/g°C * 200g * 0.70°C

Q = 585,2J

Is total heat released.

The calorimeter absorbs:

15.5J / °C * 0.7°C = 10.85

Thus, when 0.01 moles reacts, 585.2J + 10.85  = 596.05J are released (Heat released is heat abosrbed by calorimeter + Heat absorbed by water) and ΔH is:

ΔH = 596.05J / 0.01 moles =

ΔH = 59605J / mol =

<h3>ΔH = -59.6kJ/mol</h3>

<em>As heat is released, ΔH < 0.</em>

6 0
3 years ago
In a water molecule, what type of bond forms between the oxygen and hydrogen atoms
goldenfox [79]
Covalent bond forms between the oxygen and hydrogen atoms.

(I attached a picture that could help)

-Hope that helps,
Good luck!

7 0
2 years ago
Characteristics of bipolar disorder typically include
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C. Difficulty in concentrating. 
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3 0
3 years ago
Read 2 more answers
You collect 552 mL of argon gas at 23.0 C. What volume will the gas occupy at 46.0C if the pressure remains constant?
taurus [48]

Answer:

1027.9 mL

Explanation:

Formula P1 x V1 / T1 = P2 x V2 / T2

Fill in what you know

Pressure is constant so no need to put that in making the formula

V1 / T1 = V2 / T2

Voulme 1= 950 mL

Volume 2= ?

Temperature 1 = 25 C

Temperature 2 = 50 C

Explanation:

Formula P1 x V1 / T1 = P2 x V2 / T2

Fill in what you know

Pressure is constant so no need to put that in making the formula

V1 / T1 = V2 / T2

Voulme 1= 950 mL

Volume 2= ?

Temperature 1 = 25 C

Temperature 2 = 50 C

8 0
2 years ago
A solution of 1.50 g of solute dissolved in 25.0 mL of H₂O at 25°C has a boiling point of 100.45°C. (d) Find the van’t Hoff fact
givi [52]

The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

<h3>What is the value of van 't Hoff factor?</h3>

For most non-electrolytes dissolved in water, the Van 't Hoff factor is essentially $ 1 $ . For most ionic compounds dissolved in water, the Van 't Hoff factor is equal to the number of discrete ions in a formula unit of the substance.

Which has highest Van t Hoff factor?

The Van't Hoff factor will be highest for

  A. Sodium chloride.

  B. Magnesium chloride.

  C. Sodium phosphate.

  D. Urea.

To learn more about van 't Hoff factor off factor here:

brainly.com/question/22047232

#SPJ4

4 0
1 year ago
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