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3 years ago

Describe the difference between physical and chemical changes in terms of what happens to the matter

1 answer:

6 0

A physical change means that the matter will remain the same, let's say you take a piece of paper and tear it in half, it's still paper. The difference with chemical changes is, the matter changes state; now let's say you take that piece of paper and light it on fire, it's state changes to ash.

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Determine the molecular formulas of these compounds:

lianna [129]

The molecular formulas of the empirical ones of P₂O₅ and CH₂O are P₄O₁₀ and C₇H₁₄O₇, respectively, knowing that their molar masses are 310 and 200.18 g/mol, respectively.

<h3>1. Molecular formula of P₂O₅</h3>

We can calculate the molecular formula as follows:

$MF = n*EF$ (1)

Where:

MF: is the molecular formula

EF: is the empirical formula

n: is an integer

To calculate the integer n, we need to use the following equation:

$n = \frac{M}{M_{EF}}$ (2)

Where:

M: is the molar mass of P₂O₅ = 310 g/mol

$M_{EF}$ : is the molar mass of the empirical formula

The molar mass of the empirical formula of P₂O₅ is given by:

$M_{EF} = 2A_{P} + 5A_{O}$

Where:

$A_{P}$ : is the atomic weight of phosphorus = 30.974 g/mol

$A_{O}$ : is the atomic weight of oxygen = 15.999 g/mol

So, the molar mass of the empirical formula of P₂O₅ is:

$M_{EF} = (2*30.974 + 5*15.999) g/mol = 141.943 g/mol$

Now, we can find the integer n (eq 2).

$n = \frac{M}{M_{EF}} = \frac{310 \:g/mol}{141.943 \:g/mol} = 2$

Finally, after multiplying the integer n by the number of atoms on the empirical formula of P₂O₅, we have (eq 1):

$MF = n*EF = 2 \times P_{2}O_{5} = P_{(2 \times 2)}O_{(5 \times 2)} = P_{4}O_{10}$

Therefore, the molecular mass of P₂O₅ is P₄O₁₀.

<h3>2. Molecular formula of CH₂O </h3>

We know:

M: molar mass of CH₂O = 200.18 g/mol

To calculate the integer n and so the molecular mass of the molecule, we need to calculate the molar mass of the empirical formula of CH₂O.

$M_{EF} = A_{C} + 2A_{H} + A_{O} = (12.011 + 2*1.008 + 15.999) g/mol = 30.026 \:g/mol$

Now, the integer n is equal to (eq 2):

$n = \frac{M}{M_{EF}} = \frac{200.18 g/mol}{30.026 g/mol} \approx 7$

Finally, the molecular formula of the molecule is (eq 1):

$MF = n*EF = 7 \times CH_{2}O = C_{(1 \times 7)}H_{(2 \times 7)}O_{(1 \times 7)} = C_{7}H_{14}O_{7}$

Therefore, the molecular formula of CH₂O is C₇H₁₄O₇.

Learn more about molecular formula here:

brainly.com/question/1247523

brainly.com/question/14327882

I hope it helps you!

4 0

2 years ago

Which equation shows how wavelength is related to velocity and frequency?

Nataly_w [17]

Answer:

The correct option is the option;

$\lambda = \dfrac{v}{f}$

Explanation:

The wavelength of a wave is the distance between two successive crests of the wave

Therefore, the wavelength, λ, is given by the fraction of the velocity, v, of the wave divided by the frequency, f, (the number of cycles that pass through a point) of the wave

Mathematically, we have;

$The \ wavelength, \ \lambda = \dfrac{The \ wave \ velocity, \ v}{The \ wave \ frequency, \ f}$

$\lambda = \dfrac{v}{f}$

4 0

3 years ago

A resting adult requires about 240 ml of pure oxygen/mind and breathes about 12 times every minute. if inhaled air contains 20 p

Dmitry_Shevchenko [17]

12 times breathe give 240 ml of pure $O_{2}$ . Each breathe gives 20 ml of $O_{2}$ .

Let us consider, volume of air per breathe= x ml.

Pure $O_{2}$ from inhaled air= $\frac{20}{100}x$ ml and Pure $O_{2}$ from exhaled air= $\frac{16}{100}x$ ml.

Pure $O_{2}$ from inhaled and exhaled air= 20 ml

So, $\frac{20}{100}x$ + $\frac{16}{100}x$ = 20

Therefore, x = 55.5 ml

So, volume of air per breath= 55.5 ml.

7 0

3 years ago

Compare What do infrared technologies, fiber optic

LuckyWell [14K]

Answer:

ptic fiber communication and satellite communication are the leading technologies which are revolutionizing the world of telecommunications. Both technologies have their advantages and limitations which make them suitable for certain type of applications. This article will provide an overview of optic fiber and satellite communication technologies and present a comparison of the features and related issues.

Optic Fiber Communication

Optic Fiber communication transmits information by sending pulses of light (using laser) through an optic fiber. The low signal loss in optic fibers and high data rate of transmission systems, allow signals with high data rates (exceeding several Gbps) to travel over long distances (more than 100 km) without a need of repeater or amplifier. Moreover, using wavelength division multiplexing (WDM) allows a single fiber to carry multiple signals (upto 10 different signals) of multi-Gbps transmissions. Optic Fiber communication offers extremely high bandwidth, immunity to electromagnetic interference, non-existent delays and immunity from interception by external means. In the 1980s and 1990s, the continents were linked together using undersea optic fiber bringing about a paradigm shift in the global telecommunications.

These advancements in optic fiber communication has resulted in decrease of satellite communications for several types of communications. For instance, transmission between fixed locations or point-to-point communications, where large bandwidths are required (such as transoceanic telephone systems) are made through optic fiber instead of using satellite communication. Optic Fiber communication is also used to transmit telephone signals, Internet communication, LAN (Gigabit LAN) and cable television signals.

Satellite Communication

Satellite communications use artificial satellites as relays between a transmitter and a receiver at different locations on Earth. Satellite systems allow users to bypass typical carrier offices and to broadcast information to multiple locations. Communications satellites are used for radio, TV, telephone, Internet, military and other applications. There are more than 2,000 satellites around Earth’s orbit, being used for communication by both government and private organizations.

Communication Satellites are LOS (line-of-sight) microwave systems with a repeater. These satellites rotate around the earth with the speed of earth and are known as geostationary satellites. The limitations of antenna size also limits focusing capability making the coverage for a single satellite transmitter very large. This makes satellite communication ideal for TV and radio services as the signal has to flow from a single point to many points in a single direction. The large distance of satellites from the earth (about 22,300 miles) results in delays which adversely effects two-way communication like mobile conversations. Low earth orbit satellites can be used for two-way mobile communication because less power is required to reach those satellites.

Explanation:

6 0

2 years ago

Which expression could represent the

Elis [28]

There are a couple of ways in which you can express the concentration of a solution, and here they are: gram per liter (g/L), molarity (M), parts per million (ppm.), and percents (%).
As you can see, only M appears in your answers, which means that the correct option should be (2) 3.5 M.

6 0

3 years ago