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3 years ago

In the following reaction, which substance is the precipitate?(NH4)2SO4(aq) + Ba(NO3)2(aq) BaSO4(s) + 2NH4NO3(aq

2 answers:

7 0

The precipitate is the barium sulfate, or 3) BaSO4. This is because it is a solid (as seen by the (s)), unlike the other aqueous product. The fact that it is a solid means that it is insoluble in water and therefore a precipitate.

Hope this helps!

Arada [10]3 years ago

7 0

Answer: Option (3) is the correct answer.

Explanation:

A precipitate is defined as an insoluble solid formed when two aqueous solutions chemically combine to each other.

For example, $(NH_{4})_{2}SO_{4}(aq) + Ba(NO_{3})_{2}(aq) \rightarrow BaSO_{4}(s) + 2NH_{4}NO_{3}(aq)$

Since, it is known that (aq) represents aqueous and (s) represents solid. So, in this reaction barium sulphate ( $BaSO_{4}$ ) is the insoluble solid.

Hence, $BaSO_{4}$ is the precipitate that is formed.

Thus, we can conclude that in the given reaction $BaSO_{4}$ substance is the precipitate.

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1. A car travels a distance of 200 m in 4 seconds. What is the velocity of the car?

dybincka [34]

Answer:

velocity = distance / time taken

= 200/4

= 50 m/s

is the correct answer

3 0

2 years ago

A rectangular coil of 65 turns, dimensions 0.100 m by 0.200 m, and total resistance 10.0 ? rotates with angular speed 29.5 rad/s

vovikov84 [41]

Answer:

Explanation:

N = 65

Area, A = 0.1 x 0.2 = 0.02 m^2

R = 10 ohm

ω = 29.5 rad/s

B = 1 T

(a) at t = 0

e = N x B x A x ω

e = 65 x 1 x 0.02 x 29.5

e = 38.35 V

(b) The maximum rate of change of magnetic flux is equal to the maximum value of induced emf.

Ф = 38.35 Wb/s

e = 65 x 1 x 0.02 x 29.5 x Sin (29.5 x 0.05)

e = 38.174 V

(d) Maximum torque

τ = M B Sin 90

τ = N i A B

τ = N e A B / R

τ = 65 x 38.35 x 0.02 x 1 / 10

τ = 5 Nm

8 0

2 years ago

A 1500 kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2500 kg SUV traveling from ea

kolezko [41]

Answer:

v = 19.33 m / s South

Explanation:

To solve this exercise we must use the conservation of momentum, for which we must define a system formed by the two cars, therefore the forces during the collision are internal and therefore the moment is conserved.

Since it is a vector quantity, we are going to work on each axis, the x axis is in the East-West direction

initial instant. Before the crash

p₀ = m 0 + M v₂ₓ

final instant. Right after the crash

p_f = (m + M) vₓ

p₀ = 0_pf

M v₂ₓ = (m + M) vₓ

In this case m is the mass of the car and M the mass of the SUV

vₓ = $\frac{M}{m+My}$ v₂ₓ (1)

in the Y axis (North - South direction)

initial instant

p₀ = m v_{1y} + M 0

final moment

p_f = (m + M) v_y

p₀ = p_f

m v_{1y} + M 0 = (m + M) v_y

v_y = $\frac{m}{m+M} \ v_{1y}$ (2)

With these speeds we can use the relationship between work and the variation of kinetic energy, in this part the two cars are already united.

W = ΔK

friction force work is

W = - fr d

the friction force is described by the equation

fr = μ N

Newton's second law

N-W = 0

N = W

we substitute

W = - μ (m + M) g d

as the car stops the final kinetic energy is zero and

the initial kinetic energy is

K₀ = ½ (m + M) v²

we substitute

- μ (m + M) g d = 0 - ½ (m + M) v²

μ g d = ½ v²

v² = 2 μ g d

the distance traveled can be found with the Pythagorean theorem

d = $\sqrt{x^2+y^2}$

d = $\sqrt{5.48^2 + 6.37^2}$

d = 8.40 m

let's calculate the speed

v² = 2 0.75 9.8 8.40

v = √123.48

v = 11.11 m / s

this velocity is in the direction of motion so we can use trigonometry to find the angles

tan θ = y / x

θ = tan⁻¹ y / x

θ = tan⁻¹ (-5.48 / -6.37)

θ = 40.7º

Since the two magnitudes are negative, this angle is in the third quadrant, measured from the positive side of the x-axis in a counterclockwise direction.

θ'= 180 + 40.7

θ’= 220.7º

In the exercise they indicate the the sedan moves in the y-axis, therefore

sin θ'= v_y / v

v_y = v sin 220.7

v_y = 11.11 sin 220.7

v_y = -7.25 m / s

the negative sign indicates that it is moving south

To find the speed we substitute in equation 2

v_y = $\frac{m}{m+M} \ v_{1y}$