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USPshnik [31]
3 years ago
14

What are the products of the double-replacement reaction between potassium bromide and silver nitrate? Name the resulting compou

nd(s) (2 Points)
Physics
2 answers:
andreyandreev [35.5K]3 years ago
7 0
For double-replacement reaction, the cation of the first reactant would combine with the anion of the second reactant, and vice versa. Therefore, in this case, the product would be Potassium Nitrate (KNO3) and Silver Bromide (AgBr).

I hope I was able to explain it well. Have a good day :)
timofeeve [1]3 years ago
6 0

Explanation:

A double replacement reaction is a reaction in which two different compounds are mixed together and both their cations and anions get exchanged with each other respectively.

When potassium bromide reacts with silver nitrate then it results in the formation of potassium nitrate and silver bromide.

The chemical reaction equation is as follows.

        KBr + AgNO_{3} \rightarrow KNO_{3} + AgBr

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A water wave is an example of a mechanical wave. A wave that can travel only through matter is called a mechanical wave.
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2 years ago
To pull a 53 kg crate across a horizontal frictionless floor, a worker applies a force of 180 N, directed 35° above the horizont
Lorico [155]

Answer

given,

mass of the crate  = 53 Kg

force applied by the worker = 180 N

Angle made with the horizontal = 35°

crate moves = 2.9 m

a) The work done equals the force in the direction of the displacement, times the displacement

W = F_x ×d

W = 180 cos 35° × 2.9

W = 427.6 J

b) A force that is perpendicular to the direction of the displacement does not do any work so work done by gravitational force is zero

c) Similarly for the normal force the work done will be zero.

d) Total work done by the crate is equal to 427.6 J

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3 years ago
How to convert from fahrenheit to celsius
kaheart [24]
The formula for Fahrenheit and Celsius conversion is 
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8 0
3 years ago
Read 2 more answers
) A 0.10-kg ball, traveling horizontally at 25 m/s, strikes a wall and rebounds at 19 m/s. What is the magnitude of the change i
deff fn [24]

Answer:

The magnitude of the change in the momentum of the ball during the rebound is 4.4 kg-m/s.

Explanation:

Given that,

Mass of the ball, m = 0.1 kg

Initial speed of the ball, u = 25 m/s

Final speed of the ball, v = -19 m/s (the ball rebounds so it will be negative)

We need to find the magnitude of the change in the momentum of the ball during the rebound. The change in momentum of the object is equal to the difference of final and initial momentum.

\Delta p=m(v-u)

\Delta p=0.1\ kg\times (-19-25)\ m/s

\Delta p=-4.4\ kg-m/s

or

|\Delta p|=4.4\ kg-m/s

So, the magnitude of the change in the momentum of the ball during the rebound is 4.4 kg-m/s. Hence, this is the required solution.

4 0
3 years ago
A set of charged plates is
cluponka [151]

Answer:

5.49×10¯⁴ m²

Explanation:

From the question given above, the following data were obtained:

Distance (d) = 2.22×10¯⁴ m

Charge (Q) = 5.24×10¯⁹ C

Potential difference (V) = 240 V

Permittivity of free space (ε₀) = 8.85×10¯¹² F/m

Area (A) =?

Thus, the area of the plate can be obtained as follow:

Q = ε₀AV /d

5.24×10¯⁹ = 8.85×10¯¹² × A × 240 / 2.22×10¯⁴

5.24×10¯⁹ = 2.12×10¯⁹ × A / 2.22×10¯⁴

Cross multiply

2.12×10¯⁹ × A = 5.24×10¯⁹ × 2.22×10¯⁴

Divide both side by 2.12×10¯⁹

A = (5.24×10¯⁹ × 2.22×10¯⁴) / 2.12×10¯⁹

A = 5.49×10¯⁴ m²

Thus, the area of the plates is 5.49×10¯⁴ m².

8 0
2 years ago
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