How does changing the pressure affect the chemical reaction?

How do we know that caramelizing sugar is a chemical change ?

Katyanochek1 [597]

<span>Chemical transformation: it's all transformation changes the composition of matter, i.e. that matter produces a new substance with different initial properties burning sugar to make caramel, we cannot retrieve the initial sugar.
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hope this helps!

4 0

4 years ago

5. The statement that in undisturbed sedimentary rock layers,

Colt1911 [192]

Answer:

The correct answer is - option C. Law of superposition.

Explanation:

According to the law of superposition states that within a sequence of layers of natural sedimentary rock without any disturbance, the oldest layer is lies under the younger layers with ascending order in the sequence.

If a sedimentary rock has 4 layers as a, b, c and d where the younger layer d would be on top and the oldest layer a would be at the bottoms.

Thus, the correct answer is - option C.

7 0

3 years ago

Read 2 more answers

Ne, c4h10, is a component of natural gas that is used as fuel for cigarette lighters. the balanced equation of the complete comb

Simora [160]

<span>3.68 liters First, determine the number of moles of butane you have. Start with the atomic weights of the involved elements: Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Atomic weight oxygen = 15.999 Molar mass butane = 4*12.0107 + 10*1.00794 = 58.1222 g/mol Moles butane = 2.20 g / 58.1222 g/mol = 0.037851286 Looking at the balanced equation for the reaction which is 2 C4H10(g)+13 O2(g)â†’8 CO2(g)+10 H2O(l) It indicates that for every 2 moles of butane used, 8 moles of carbon dioxide is produced. Simplified, for each mole of butane, 4 moles of CO2 are produced. So let's calculate how many moles of CO2 we have: 0.037851286 mol * 4 = 0.151405143 mol The ideal gas law is PV = nRT where P = Pressure V = Volume n = number of moles R = Ideal gas constant ( 0.082057338 L*atm/(K*mol) ) T = absolute temperature (23C + 273.15K = 296.15K) So let's solve the formula for V and the calculate using known values: PV = nRT V = nRT/P V = (0.151405143 mol) (0.082057338 L*atm/(K*mol))(296.15K)/(1 atm) V = (3.679338871 L*atm)/(1 atm) V = 3.679338871 L So the volume of CO2 produced will occupy 3.68 liters.</span>

4 0

3 years ago

You want to prepare 500.0 mL of 1.000 M KNO3 at 20°C, but the lab (and water) temperature is 24°C at the time of preparation. Ho

timama [110]

Explanation:

As per the given data, at a higher temperature, at $24^{o}C$ , the solution will occupy a larger volume than at $20^{o}C$ .

Since, density is mass divided by volume and it will decrease at higher temperature.

Also, concentration is number of moles divided by volume and it decreases at higher temperature.

At $20^{o}C$ , density of water=0.9982071 g/ml

Therefore, $\frac{concentration}{density}$ will be calculated as follows.

= $\frac{C_{1}}{d_{1}}$

= $\frac{1.000 mol/L}{0.9982071 g/ml}$

= 1.0017961 mol/g

At $24^{o}C$ , density of water = 0.9972995 g/ml

Since, $\frac{concentration}{density}$ = $\frac{C_{2}}{d_{2}}$

= $\frac{C_{2}}{0.9972995}$

Also, $\frac{C_{1}}{d_{1}}$ = $\frac{C_{2}}{d_{2}}$

so, 1.0017961 mol/g = $\frac{C_{2}}{0.9972995}$

$C_{2} = 1.0017961 \times 0.9972995$

= 0.9990907 mol/L

Therefore, in 500 ml, concentration of $KNO_{3}$ present is calculated as follows.

$C_{2}$ = $\frac{concentration}{volume}$

0.9990907 mol/L = $\frac{concentration}{0.5 L}$

concentration = 0.49954537 mol

Hence, mass (m'') = $0.49954537 mol \times 101.1032 g/mol$ = 50.5056 g (as molar mass of $KNO_{3}$ = 101.1032 g/mol).

Any object displaces air, so the apparent mass is somewhat reduced, which requires buoyancy correction.

Hence, using Buoyancy correction as follows,

m = $m''' \times \frac{(1 - \frac{d_{air}}{d_{weights}})}{(1 - \frac{d_{air}}{d})}}$

where, $d_{air}$ = density of air = 0.0012 g/ml

$d_{weight}$ = density of callibration weights = 8.0g/ml

d = density of weighed object

Hence, the true mass will be calculated as follows.

True mass(m) = $50.5056 \times \frac{(1 - \frac{0.0012}{8.0})}{(1 - (\frac{0.0012}{2.109})}$

true mass(m) = 50.5268 g

= 50.53 g (approx)

Thus, we can conclude that 50.53 g apparent mass of $KNO_{3}$ needs to be measured.

8 0

3 years ago

What two-step process separated the cans into aluminum, steel, and tin?

Vikentia [17]

Answer:

Magnet

Durability and heaviness.(texture)

Explanation:

Magnet can be use to separate Aluminum from mixture of steel and aluminum.

Though aluminum and steel look alike but magnet can be use to separate it.

If the can attract the magnet or magnet stick to the can, it is a steel can. Aluminum does not stick to magnet.

A mixture of Aluminum and tin can also be separated by magnet.

Tin attract magnet but tin is more durable, heavy and does not corrode easily.

When u touch the three cans, tin is heavy and durable.

4 0

4 years ago