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sineoko [7]
3 years ago
14

The caloric content of 0.50 cup of cottage cheese is 1.10 x 102 kcal. This is the amount of energy released when 0.50 cup of cot

tage cheese is burned or metabolized by the body. The average slow walker requires 226 kcal of energy to walk one mile. How many cups of cottage cheese would provide enough energy for this walker to walk from Gannon’s Zurn Science Center to the Millcreek Mall (19.3 km)?
Chemistry
1 answer:
V125BC [204]3 years ago
3 0

Answer:

A piece of unknown solid substance weighs 437.2 g, and requires 8460 J to increase its temperature from 19.3 °C to 68.9 °C.

What is the specific heat of the substance?

If it is one of the substances found in Table 8.1.1, what is its likely identity?

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Use the words protons neutrons and I suppose in the same sentence ​
PolarNik [594]

Answer:

Isotopes have the same number of protons but a different number of neutrons.

Explanation:

5 0
3 years ago
Why is the energy supplied by the cooker greater than that calculated ?
TEA [102]

Answer:

Explanation:

Q1.

(a) 46 200

accept 46 000

allow 1 mark for correct substitution

ie 0.5 × 4200 × 22 provided no subsequent step

2

(b) Energy is used to heat the kettle.

[3]

Q2.

(a) (approximate same size particles as each other and as liquid and gas) touching

do not accept particles that overlap

regular arrangement (filling the square)

(b) condensing

(c) solid

(d) physical

(e) particles have more kinetic energy

particles move faster

(f) mass of the liquid

specific latent heat of evaporation

(g) 2 × 4 200 × 801

672 000 (J)

an answer of 672 000 (J) scores 2 marks

[11]

Q3.

(a) x-axis labelled and suitable scale

Page 12 of 13

points plotted correctly

allow 5 correctly plotted for 2 marks, 3−4

correctly plotted for 1 mark

allow ± ½ square

2

line of best fit

(b)

allow ecf from line of best fit in part (a)

0.075 (°C/s)

an answer of 0.075 (°C/s) scores 2 marks

(c) Δθ = 11.5 (°C)

a calculation using an incorrect temperature

scores max 3 marks

ΔE = 1.50 × 900 × 11.5

ΔE = 15 525 (J)

ΔE = 15.525 (kJ)

an answer of 15.525 (kJ) or 15.53 (kJ) or 15.5

(kJ) scores 4 marks

an answer of 15 525 (kJ) scores 3 marks

[10]

Q4.

(a) 80 °C

ΔE = 0.5 × 3400 × 80

ΔE = 136 000 (J)

an answer of 136 000 (J) scores 3 marks

(b) energy is dissipated into the surroundings

allow any correct description of wasted energy

(c) put a lid on the pan

allow any sensible practical suggestion

eg add salt to the water

Page 13 of 13

(d) efficiency = 300/500

efficiency = 0.6

an answer of 0.6 or 60% scores 2 marks

allow efficiency = 60%

an answer of 0.6 with a unit scores 1 mark

an answer of 60 without a unit scores 1 mark

7 0
2 years ago
having that coloration is selected for since many predators will recognize it and stay away. Therefore the common coloration is
garri49 [273]

Answer: Natural selection.

Explanation:

The natural selection is the process by which the organism that best suits the characters will survive and the organism which cannot adapt the changes will die.

The predators feed on their prey and the organism which hides easily in the environment due to its color is safe and the animals that cannot camouflage will die.

This is natural selection that more favored character is carried out over generations and other that cannot survive die.

5 0
3 years ago
Calculate the molar solubility of CaF2 in a 0.25 m solution of NaF(aq).
Kipish [7]

Answer:

6.4 × 10^-10 M

Explanation:

The molar solubility of the ions in a compound can be calculated from the Ksp (solubility constant).

CaF2 will dissociate as follows:

CaF2 ⇌Ca2+ + 2F-

1 mole of Calcium ion (x)

2 moles of fluorine ion (2x)

NaF will also dissociate as follows:

NaF ⇌ Na+ + F-

Where Na+ = 0.25M

F- = 0.25M

The total concentration of fluoride ion in the solution is (2x + 0.25M), however, due to common ion effect i.e. 2x<0.25, 2x can be neglected. This means that concentration of fluoride ion will be 0.25M

Ksp = {Ca2+}{F-}^2

Ksp = {x}{0.25}^2

4.0 × 10^-11 = 0.25^2 × x

4.0 × 10^-11 = 0.0625x

x = 4.0 × 10^-11 ÷ 6.25 × 10^-2

x = 4/6.25 × 10^ (-11+2)

x = 0.64 × 10^-9

x = 6.4 × 10^-10

Therefore, the molar solubility of CaF2 in NaF solution is 6.4 × 10^-10M

8 0
3 years ago
How are the following aspects of a reaction affected by the addition of a catalyst? 1) activation energy of the reverse reaction
snow_tiger [21]

These are four questions and four answers.

Answers:

1) activation energy of the reverse reaction

     b. Decreased

2) Rate of the forward reaction

    a. Increased

3) Rate of the reverse reaction

    a. Increased

4) Activation energy of the forward reaction

    b. decreased

Explanation:

<em>Activarion energy</em> is the energy required by the reactants to form the intermediate transition state and become products.

<em>Catalysts</em> are substances that change the path of the chemical reactions, lowering the activation energy, and thus speeding up the rate of the reactions, since the products can reach the new lower activation energy faster.

The equilibrium reactions are the chemical process in which two reactions, the <em>forward and the reverese reactions</em>, occur simultaneously and at the same rate.  The equlibrium reactions may be represented by:

  • A ⇄ B

Where A → B is the direct or forward reaction, and A ← B is the reverse reaction (note the inversed arrow, from right to left).

For the direct reaction A represents the reactants and B represents the products. On the other hand, B represents the reactants and A represents the reactants of the reverse reaction and A. This, is A is the reactant of the forward reaction and the product of the reverse reaction, while B is the reactant of the reverse reaction and the product of the forward reaction.

Since, <em>the addition of a catalyst</em> lowers the activation energy of the process, the new activation energy is lower for both the forward and the reverse reaction, meaning that:

1. <em>The activation energy of the reverse reaction is decreased</em> (option b. of the first question)

2.<em> The rate of the forward reaction is increased</em> (option a. of the second question)

3. <em>The rate of the reverse reaction is increased</em> (option a. of the third question).

4. <em>Activation energy of the forward reaction is decreased</em> (option b. of the fourth question).

In summary, the addition of a catalyst decreases the activation energy for both forward and reverse reactions, and increases the rate of both forward and reverse reactions.

3 0
3 years ago
Read 2 more answers
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