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3 years ago

How many total atoms are in 0.250 g of p2o5?

2 answers:

5 0

There are 2 P atoms and 5 O atoms which results in a total of 7 atoms per molecule. Hope this helps you out; I had the same question yesterday on my 9th grade Physical Science test.

Strike441 [17]3 years ago

3 0

Answer:

3.71 × 10²¹ atoms

Explanation:

The molar mass of P₂O₅ is 283.89 g/mol. The moles of P₂O₅ corresponding to 0.250 g of P₂O₅ are:

0.250 g × (1 mol/283.89 g) = 8.81 × 10⁻⁴ mol

In 1 mole of P₂O₅ there are 6.02 × 10²³ molecules of P₂O₅ (Avogadro's number). In 8.81 × 10⁻⁴ moles of P₂O₅, the molecules of P₂O₅ are:

8.81 × 10⁻⁴ mol × (6.02 × 10²³ molecule/ 1 mol) = 5.30 × 10²⁰ molecule

In 1 molecule of P₂O₅, there are 7 atoms (2 atoms of P and 5 atoms of O). In 5.30 × 10²⁰ molecules of P₂O₅, the number of atoms is:

5.30 × 10²⁰ molecule × (7 atom/1 molecule) = 3.71 × 10²¹ atom

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If a cannonball is fired horizontally it will not go in a straight line why?

goldfiish [28.3K]

Explanation:

A projectile (Cannon ball) is launched at an angle to the horizontal and rises up to a peak while moving horizontally. When it reaches the peak, the projectile starts to fall.

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1 year ago

A motorboat maintained a constant speed of 15 miles per hour relative to the water in going 10 miles upstream and then returning

Lesechka [4]

Answer:

speed of current is 5 mile/hr

Explanation:

GIVEN DATA:

speed of motorboat = 15 miles/hr relative with water

let c is speed of current

15-c is speed of boat at upstream

15+c is speed of boat at downstream

we know that

travel time=distance/speed

$\frac{10}{15-c} +\frac{10}{15+c} = 1.5$

150+10c+150-10c=1.5(15-c)(15+c)

300=1.5(225-c^2)

300=337.5-1.5c^2

200=225-c^2

c^2=25

c = 5

so speed of current is 5 mile/hr

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2 years ago

This chemical equation represents the burning of methane, but the equation is incomplete. What is the missing coefficient in bot

kolezko [41]

Your complete chemical equation is CH4+O2=CO3+H2O

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3 years ago

Read 2 more answers

An apple dropped from the branch of a tree hits the ground in 0.5 s. If the acceleration of the apple during its motion is 10 ms

Ipatiy [6.2K]

Given that,

Time = 0.5 s

Acceleration = 10 m/s²

(I). We need to calculate the speed of apple

Using equation of motion

$v=u+at$

Where, v = speed

u = initial speed

a = acceleration

t = time

Put the value into the formula

$v=0+10\times0.5$

$v=5\ m/s$

(III). We need to calculate the height of the branch of the tree from the ground

Using equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Put the value into the formula

$s=0+\dfrac{1}{2}\times10\times(0.5)^2$

$s=1.25\ m$

(II). We need to calculate the average velocity during 0.5 sec

Using formula of average velocity

$v_{avg}=\dfrac{\Delta x}{\Delta t}$

$v_{avg}=\dfrac{x_{f}-x_{i}}{t_{f}-t_{0}}$

Where, $x_{f}$ = final position

$x_{i}$ = initial position

Put the value into the formula

$v_{avg}=\dfrac{1.25+0}{0.5}$

$v_{avg}=2.5\ m/s$

Hence, (I). The speed of apple is 5 m/s.

(II). The average velocity during 0.5 sec is 2.5 m/s

(III). The height of the branch of the tree from the ground is 1.25 m.

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2 years ago

Suppose you had the same laser and diffraction grating from the previous question but now you had a flat detection screen. You w

kolezko [41]

Answer:

measuring the zero intensity point, we can deduce the movement of the screen.

The distance from the center of the pattern to the first zero is proportional to the distance to the screen,

Explanation:

The expression for the diffraction phenomenon is

a sin θ = m λ

for the case of destructive interference. In general the detection screen is quite far from the grid, let's use trigonometry to find the angles

tan θ = y / L

in these experiments the angles are small

tan θ = sin θ / cos θ = sin θ

sunt θ = y / L

we substitute

a $\frac{y}{L}$ = m λ

y = m L λ / a

therefore, by carefully measuring the zero intensity point, we can deduce the movement of the screen.

The distance from the center of the pattern to the first zero is proportional to the distance to the screen, so you can know where the displacement occurs, it should be clarified that these displacements are very small so the measurement system must be capable To measure quantities on the order of hundredths of a millimeter, a micrometer screw could be used.

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2 years ago