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maks197457 [2]
3 years ago
7

Is anyone good at chemistry if so can someone help me please ? (NO LINKS)

Chemistry
1 answer:
Mkey [24]3 years ago
6 0

This requires familiarity with the different theories (or concepts) of acids and bases.

On the Arrhenius concept, an acid is a substance that produces an H⁺ ion in water such that the H⁺ concentration increases, and a base is a substance that produces an OH⁻ ion in water such that the OH⁻ concentration increases.

On the Brønsted–Lowry concept, an acid is a substance that donates a proton (which is basically an H⁺ ion) in a solvent, and a base is a substance that accepts a proton in a solvent.

On the Lewis concept, an acid is a substance that accepts an electron pair in a solvent, and a base is a substance that donates an electron pair in a solvent.

The concepts become progressively broader, i.e., the Arrhenius concept is the most restrictive and the Lewis concept is the least restrictive. As a corollary, an Arrhenius acid or base is also both a Brønsted–Lowry acid or base and a Lewis acid or base, respectively; a Brønsted–Lowry acid or base is not necessarily an Arrhenius acid or base, but an Arrhenius acid or base is also a Lewis acid or base, respectively. And finally, a Lewis acid or base may not necessarily be either an Arrhenius or a Brønsted–Lowry acid or base.

So, with the above concepts in mind, we can match the statements in column A with the type of acid or base in column B:

\begin{center}\begin{tabular}{ c c } 1 & Bronsted Lowry acid \\  2 & Bronsted Lowry base \\   3 & Arrhenius acid \\ 4 & Arrhenius base \\ 5 & Lewis base \\ 6 & Lewis acid\end{tabular}\end{center}

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if a solution measured with the miscalibrated meter (above) gave a reading of 6.50 at 20oC, what should be the true pH (give 2 d
Andreas93 [3]

Answer:

Explanation:

Ph is the major of acidity or basicity of a solution. On a scale of 0 to 14

PH = -log 10 ^(H+)

Therefore PH= -log 10^(6.5) =0.81

Therefore PH of the solution is 0.81

This solution is therefore an acidic solution

7 0
4 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)

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(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

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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

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allow any sensible practical suggestion

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(d) efficiency = 300/500

efficiency = 0.6

an answer of 0.6 or 60% scores 2 marks

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7 0
3 years ago
What is the change in enthalpy when 11. 0 g of liquid mercury is heated by 15°c?
Vlad1618 [11]

Enthalpy change is the difference between energy used and energy gained. The change in enthalpy of the liquid mercury is 0.0231 kJ.

<h3>What is the enthalpy change?</h3>

Enthalpy change is the difference between the energy used to break chemical bonds and the energy gained by the products formed in a chemical reaction.

The enthalpy change is given by,

\rm \Delta H_{rxn} = \rm q_{rxn}

and,

\rm q = mc\Delta T

Given,

Mass of the liquid mercury (m) = 11.0 gm

The specific heat of mercury (c) = 0.14 J per g per degree Celsius

Temperature change = 15 degrees Celsius

Enthalpy change is calculated as:

\begin{aligned} \rm q &= \rm mc\Delta T\\\\&= 11 \times 0.14 \times 15\\\\&= 23.1 \;\rm J\end{aligned}

Therefore, 0.0231 kJ is the change in enthalpy.

Learn more about enthalpy change here:

brainly.com/question/10932978

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6 0
2 years ago
How is data not actually obtained from the experiment represented in a line graph?
zavuch27 [327]
A colored line, as long as it is one single piece, not broken
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3 years ago
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A nugget of GOLD has a mass 9.66 gram and a volume of 0.5 cm3, It's density is<br>grams/cm3 *​
dalvyx [7]

Answer:

P=19.32g/cm³

Explanation:

m=9.66g

v=0.5cm³

P=mass/volume (density formula)

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=19.32g/cm³

4 0
3 years ago
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