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3 years ago

Examine the reactants of the incomplete double displacement reaction.

Physics

1 answer:

Bumek [7]3 years ago

8 0

<h3>Answer:</h3>

AgCl + NaNO₃

<h3>Explanation:</h3>

The reaction between silver nitrate and sodium chloride is an example of a double displacement reaction.
In a double displacement reaction compounds or salts reacts and exchange cations or anions to form new compounds or salts.
In this case, silver nitrate and sodium chloride exchange anions and cations to form silver chloride and sodium nitrate.

Therefore, the complete reaction is given by;

AgNO3 + NaCl → AgCl + NaNO₃

But since silver chloride is a precipitate, the reaction may also be an example of a precipitation reaction.

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4 years ago

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Nikolay [14]

Answer:

Explanation:

The magnitude of the electrostatic force between two charged particles is given by

$F=\frac{kq_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the charges of the two particles

r is the separation between the particles

In this problem, the initial force between the particles is F.

Later, the distance between the two particles is increased by four, so

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So, the new force between the particles will be

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3 years ago

Which solutions have a density greater than 1.0 g/mL?

fenix001 [56]

Density = Mass / Volume

A) Density = 888/800 = 1.11g/mL > 1.0g/mL
B) Density = 7.8 / 8.7 = 0.897g/mL < 1.0g/mL
C) density = 725 / 715 = 1.01 g/mL > 1.0g/mL
D)density = 1.3/1.1= 1.18 g/mL > than 1.0 g/mL
E) Density = 18/19 = 0.95g/mL < 1.0g/mL
F) Density = 1.25/1.78 = 0.7 g/mL < 1.0g/mL

SUMMARY:
A C D have densities greater than 1.0g/mL

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3 years ago

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4 years ago

A rifle is aimed horizontally at a target 49 m away. the bullet hits the target 2.3 cm below the aim point. part a what was the

Nezavi [6.7K]

We use the equation of motion for vertical component,

$s_{y} = u_{y} t+\frac{1}{2} gt^2.$

Here, $s_{y}$ is displacement of bullet, $u_{y}$ is vertical initial velocity of bullet which is equal to zero because bullet was fired horizontally, and t is time of flight.

Therefore,

$s_{y} =\frac{1}{2}gt^2$

Given, $s_{y} =2.3 \ cm = 2.3 \times 10^{-2} \ m$

Substituting the values, we get time of flight

$2.3 \times 10^{-2} \ m = \frac{1}{2} \times 9.8 \ m/s^2 \times t^2 \\\\ t =\sqrt{46.94 \times 10^{-4} \ s } = 0.069 \ s$

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