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3 years ago

A compunds whose empirical formula is xf3 consists of 65% f by mass. what is the atomic mass of x

Physics

2 answers:

pogonyaev3 years ago

5 0

The atomic mass of X is 30.69 u.

<h3>Further Explanation</h3>

Mass percent is the ratio of an element's mass to the mass of the compound. It is mathematically expressed as:

$\% \ mass \ of \ A \ = \frac{mass \ of \ element \ A}{mass \ of \ compound} \times 100$

In the problem, the mass percent of F is given and can be used to determine the atomic mass of element X. To do this, just set up the mass percent equation for F as shown below:

$mass \ \% \ of \ F = \frac{3 \times 18.998 \ u}{ X + 56.994 u}\\\\mass \% \ of \ F = \frac{56.994 \ u}{X + 56.994 \ u} = 65\%\\$

Solving for X,

$0.65 (X + 56.994) = 56.994\\0.65X \ + \ 37.0461 = 56.994\\0.65X = 19.9479\\\\\boxed {\boxed {X = 30.69 \ u}}$

To check if the atomic mass of X is correct, use this value to calculate the %F. It is is equal to 65%, then the calculated value for X is correct.

$\% F \ = \frac{3 \times 18.998 u}{30.69u \ + 56.994 u} \times 100\\\\\%F \ = \frac{56.994u}{87.684u} \times 100\\\\\\boxed {\%F = 64.999\%}$

Since substituting 30.69 as the atomic mass of X gives a mass % of 65 for F, then the calculated atomic mass is correct.

<h3>Learn More</h3>

Learn more about empirical formula brainly.com/question/8516072
Learn more about chemical formula brainly.com/question/4697698
Learn more about percent composition brainly.com/question/11042307

Keywords: mass percent, empirical formula

Reika [66]3 years ago

4 0

Empirical formula of compound is XF3

Compound consist of 65% F

In 100g of compound there is 65 g of F

= 65 / 19 moles of Fluorine = 3.421 moles

So moles of X = 3.421 / 3 = 1.140 moles

And in 100 g X consist of 35 g

So the molar mass of X = 35 / 1.140 = 30.71 g = 31 approximately

And it is the mass of phosphorus

So the empirical formula for the compound is PX3

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pogonyaev

Number of miles that marker shows when passes through town= 160 miles.

Number of miles that marker shows currently to John = 115 miles.

We need to find the distance between town and John's current location.

For the problem, we can clearly see that Town is at 160 miles away but when John passes the marker shows 115 miles.

So, it's just the difference between 160 miles and 115 miles.

In order to find that difference, we need to subtract those two numbers.

160miles - 115miles = 45 miles.

So, we could say the distance between town and John's current location is 45 miles.

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3 years ago

The international Space Station (ISS) orbits the Earth once every 90 mins at an altitude of 409 km. How high would it have to be

Oksi-84 [34.3K]

It would have to be 36,719 Km high in order to be to be in geosynchronous orbit.

To find the answer, we need to know about the third law of Kepler.

<h3>What's the Kepler's third law?</h3>

It states that the square of the time period of orbiting planet or satellite is directly proportional to the cube of the radius of the orbit.
Mathematically, T²∝a³

<h3>What's the radius of geosynchronous orbit, if the time period and altitude of ISS are 90 minutes and 409 km respectively?</h3>

The time period of geosynchronous orbit is 24 hours or 1440 minutes.
As the Earth's radius is 6371 Km, so radius of the ISS orbit= 6371km + 409 km = 6780km.
If T1 and T2 are time period of geosynchronous orbit and ISS orbit respectively, a1 and a2 are radius of geosynchronous orbit and ISS orbit, as per third law of Kepler, (T1/T2)² = (a1/a2)³
a1= (T1/T2)⅔×a2

= (1440/90)⅔×6780

= 43,090 km

Altitude of geosynchronous orbit = 43,090 - 6371= 36,719 km

Thus, we can conclude that the altitude of geosynchronous orbit is 36,719km.

Learn more about the Kepler's third law here:

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2 years ago

A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.

AveGali [126]

Answer:

The magnitude of electric force is $7.2\times10^{-3} N$

Explanation:

Coulomb's Law:

The force of attraction or repletion is

directly proportional to the products of charges i.e $F\propto q_1q_2$

inversely proportional to the square of distance i.e $F\propto \frac{1}{r^2}$

$\therefore F\propto \frac{q_1q_2}{r^2}$

$\Rightarrow F=k \frac{q_1q_2}{r^2}$ [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, $r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5$

we know,

$cos \theta = \frac{base }{hypotenuse}$

$\Rightarrow cos \theta = \frac{4 }{r}$

Total force $F_Q = 2.F_1 cos\theta \hat{i}$

$=2k\frac{Qq_1}{r^2} cos\theta \hat i$

$=2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i$

$=8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i$ [ r=5]

$=7.2\times10^{-3}\hat i$ N

The magnitude of electric force is $7.2\times10^{-3} N$

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Between magnitude of the average 4sec

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Fynjy0 [20]

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Blowout skid and

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