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4 years ago

How Do We Know That Caramelizing Sugar Is A Chemical Change?

Chemistry

2 answers:

zavuch27 [327]4 years ago

8 0

the correct answer is d. it changes the sugars color , a. is a physical change .

34kurt4 years ago

7 0

The answer is B. Sugar is made of carbon, hydrogen,and oxygen atoms. When heated, these elements react to the fire to turn into a liquid. The heat causes the sugar's atoms to combine with the oxygen in the air, forming new groups of atoms. Energy is released in the form of smoke. Let me know if it's right :)

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Determine the energy of 1.70 mol of photons for each of the following kinds of light. (Assume three significant figures.)PART A

BabaBlast [244]

Answer:

For A: The energy of the given amount of photons for infrared radiation is $1.271\times 10^5J$

For B: The energy of the given amount of photons for infrared radiation is $4.026\times 10^5J$

For C: The energy of the given amount of photons for infrared radiation is $1.355\times 10^6J$

Explanation:

The relationship between energy and frequency is given by Planck's equation, which is:

$E=n\rimes N_A\times \frac{hc}{\lambda}$ ......(1)

where,

h = Planck's constant = $6.62\times 10^{-34}Js$

E = energy of the light

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength of light

$N_A$ = Avogadro's number = $6.022\times 10^{23}$

n = number of moles of photons = 1.70 moles

Conversion factor used: $1m=10^9nm$

For A:

Wavelength of infrared radiation = $1600nm=1.6\times 10^6m$

Putting values in equation 1, we get:

$E=1.7\times 6.022\times 10^{23}\times \frac{6.62\times 10^{-34}\times 3\times 10^8}{1.6\times 10^{-6}}\\\\E=1.271\times 10^5J$

Hence, the energy of the given amount of photons for infrared radiation is $1.271\times 10^5J$

For B:

Wavelength of visible light = $505nm=5.05\times 10^7m$

Putting values in equation 1, we get:

$E=1.7\times 6.022\times 10^{23}\times \frac{6.62\times 10^{-34}\times 3\times 10^8}{5.05\times 10^{-7}}\\\\E=4.026\times 10^5J$

Hence, the energy of the given amount of photons for infrared radiation is $4.026\times 10^5J$

For C:

Wavelength of ultraviolet radiation = $150nm=1.5\times 10^7m$

Putting values in equation 1, we get:

$E=1.7\times 6.022\times 10^{23}\times \frac{6.62\times 10^{-34}\times 3\times 10^8}{1.5\times 10^{-7}}\\\\E=1.355\times 10^6J$

Hence, the energy of the given amount of photons for infrared radiation is $1.355\times 10^6J$

3 0

4 years ago

Predict the products of the following reaction and balance the equation: ___Na2CO3(aq) + ___HCl (aq) ->

zheka24 [161]

Answer:

RxN: Na₂CO₃ (aq) + 2HCl (aq) → 2NaCl (aq) + H₂CO₃ (aq)

NIE: n/a

Explanation:

Step 1: Predict (Double Replacement)

Na₂CO₃ (aq) + HCl (aq) → NaCl (aq) + H₂CO₃ (aq)

Step 2: Balance

Na₂CO₃ (aq) + 2HCl (aq) → 2NaCl (aq) + H₂CO₃ (aq)

Step 3: Write Total Ionic Equation

2Na⁺ (aq) + CO₃²⁻ (aq) + 2H⁺ (aq) + 2Cl⁻ (aq) → 2Na⁺ (aq) + 2Cl⁻ (aq) + 2H⁺ (aq) + CO₃²⁻ (aq)

Step 4: Remove Spectator Ions

- Na⁺

- CO₃²⁻

- H⁺

- Cl⁻

Step 5: Write Net Ionic Equation

None.

3 0

3 years ago

The main purpose of the turbine in the turbojet engine is to A. drive the compressor. B. increase the velocity of the exhaust ga

Naddika [18.5K]

The main purpose of the turbine in the turbojet engine is to drive the compressor.

Answer: A) or the first option.

7 0

4 years ago

Read 2 more answers

How is the polar vortex different from the jet stream?

kolezko [41]

The jet stream is a band of reliably strong wind that plays a key role in keeping colder air north and warmer air south. But when the vortex weakens, part of the weakened low-pressure system can break off. This breaking-off process is what causes a polar vortex.

Without that strong low-pressure system, the jet stream does not have enough force to maintain its usual path. It becomes wavy and rambling. When high-pressure systems get in its way, a collection of cold air pushes south, along with the rest of the polar vortex system

8 0

3 years ago

According to Graham’s law, the rate of effusion of a gas is inversely proportional to

VLD [36.1K]

C. the square root of the mass of the particles.

<h3>Further explanation </h3>

Graham's law: the rate of effusion of a gas is inversely proportional to the square root of its molar masses or

the effusion rates of two gases = the square root of the inverse of their molar masses:

$\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }$

$\rm M_1\times r_1^2=M_2\times r_2^2$

From this equation shows that the greater the mass of the gas, the smaller the effusion rate of the gas and vice versa, the smaller the mass of the gas, the greater the effusion velocity.

So if both gases are at the same temperature and pressure, the above formula can apply

5 0

3 years ago