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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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What two things must all matter have?
OleMash [197]

1.) Mass

2.) Can occupy space (Volume)

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If you filter a mixture of sand and salt-water, what will the solid residue in the filter paper consist of?
julia-pushkina [17]
Sand only bc the salt dissolved and sand doesn't
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Does a precipitate form when a solution of calcium chloride and a solution of mercury(I) nitrate are mixed together? Write the n
olganol [36]

Answer : Yes, a precipitate form when a solution of calcium chloride and a solution of mercury(I) nitrate are mixed together.

The net ionic equation will be,

2Hg^{+}(aq)+2Cl^{-}(aq)\rightarrow Hg_2Cl_2(s)

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The given balanced ionic equation will be,

CaCl_2(aq)+2HgNO_3(aq)\rightarrow Ca(NO_3)_2(aq)+Hg_2Cl_2(s)

The ionic equation in separated aqueous solution will be,

Ca^{2+}(aq)+2Cl^{-}(aq)+2Hg^{+}(aq)+2NO_3^{-}(aq)\rightarrow Hg_2Cl_2(s)+Ca^{2+}(aq)+2NO_3^{-}(aq)

In this equation, Ca^{2+}\text{ and }NO_3^- are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

2Hg^{+}(aq)+2Cl^{-}(aq)\rightarrow Hg_2Cl_2(s)

7 0
3 years ago
A plot of binding energy per nucleon (Eb/ A) versus the mass number (A) shows that nuclei with a small mass number have a small
juin [17]

Answer:

a)   1.12 MeV / nucleon

b)   5.62 MeV / nucleon

c)  8.80 MeV / nucleon

d) 8.56 MeV / nucleon

we can conclude that the binding energy has a maximum value for nuclei with a mass around 60

Explanation:

Binding energy = ( Δm * 931.5 ) MeV

Binding energy per nucleon = Binding energy in / Number of nucleon

<u>a) ²H = 1 neutron , 1 proton = 2 nucleons </u>

Given that the theoretical mass = 2.0141 u

Actual mass = 1.0078 u + 1.0087 u = 2.0165 u

Δm  = 2.0165 u - 2.0141 u = 2.4 * 10^-3 u

∴ Binding energy per nucleon = ( 2.4 * 10^-3  * 931.5 ) MeV / 2 nucleons

                                                  = 1.12 MeV / nucleon

<u>b) ⁷Li = 3 protons , 4 neutrons = 7 nucleons </u>

theoretical mass = 7.0160 u

Actual mass = ( 3 * 1.0078 ) + ( 4 * 1.0087 )  = 7.0582 u  

Δm  = ( 7.0582 u  - 7.0160 u  ) = 0.0422 u

∴ Binding energy per nucleon = ( 0.0422 * 931.5 ) / 7

                                                  = 5.62 MeV / nucleon

<u>C) ⁶²Ni = 28 protons , 34 neutrons = 62 nucleons </u>

Theoretical mass = 61.9283 u

Actual mass = ( 28 * 1.0078 ) u + ( 34 * 1.0087 ) u

                    = 62.5142 u

Δm = 0.5859 u  

∴ Binding energy per nucleon = ( 0.5859 * 931.5 ) / 62

                                                  = 8.80 MeV / nucleon

<u>D) ¹¹⁰Cd = 48 protons , 62 neutrons = 110 nucleons </u>

Theoretical mass = 109.9030 u

Actual mass = ( 48 * 1.0078 ) + ( 62 * 1.0087 )

                    = 110.9138 u

Δm  = ( 110.9138 - 109.9030 ) = 1.0108 u

∴ Binding energy per nucleon = ( 1.0108 * 931.5 ) / 110

                                                  = 8.56 MeV / nucleon

hence we can conclude that the binding energy has a maximum value for nuclei with a mass around 60

3 0
3 years ago
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