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katovenus [111]
3 years ago
7

Which food is mainly made up of carbohydrates?

Chemistry
1 answer:
Alenkasestr [34]3 years ago
6 0
Bread is the answer you are looking for please mark as brainliest! :)
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Does the energy go from the surrounding to the chemicals or from the chemicals to the surround in an exothermic reaction?
Karo-lina-s [1.5K]
In general terms, exothermic reactions release energy, so the energy goes from the system to the surroundings.
4 0
3 years ago
What is the future for the spectroscope?
boyakko [2]

Answer:

more people may make more changes to it.

there is the answer to it!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

5 0
3 years ago
Read 2 more answers
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
A sample of 42 mL of carbon dioxide gas was placed in a piston in order to maintain a constant 101 kPa of pressure.
Lilit [14]

Answer:

The answer to your question is 33.4 ml

Explanation:

Data

volume 1 = V1 = 42 ml

temperature 1 = T1 = 20°C

temperature 2 = T2 = -60°C

Volume 2 = V2 = x

Process

1.- Convert celsius to kelvin

T1 = 20 + 273 = 293°K

T2 = -60 + 273 = 233°K

2.- Use the Charles' law to solve this problem

               \frac{V1}{T1} = \frac{V2}{T2}

Solve for V2

                V2 = \frac{V1T2}{T1}

3.- Substitution

               V2 = \frac{(42)(233)}{293}

4.- Simplification

                V2 = \frac{9786}{293}

5.- Result

                V2 = 33.4ml

3 0
3 years ago
Given the following data:
bagirrra123 [75]

176.0 \; \text{kJ} \cdot \text{mol}^{-1}

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its \Delta H can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with \Delta H given be denoted as (1), (2), (3), and the last equation (4). Let a, b, and c be letters such that a \times (1) + b \times (2) + c \times (3) = (4). This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, 3 + (-1) = 2 shall resemble the number of \text{H}_2 left on the product side when the second equation is directly added to the third. Similarly

  • \text{NH}_4 \text{Cl} \; (s): -2 \; a = 1
  • \text{NH}_3\; (g): -2 \; b = -1
  • \text{HCl} \; (g): 2 \; c = -1

Thus

a = -1/2\\b = 1/2\\c = -1/2 and

-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)

Verify this conclusion against a fourth species involved- \text{N}_2 \; (g) for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

a + b = -1/2 + 1/2 = 0

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}

3 0
3 years ago
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