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2 years ago

9

The reactants for a certain chemical reaction are shown. Cu(NO3)2 + NaOH → Which of the following shows the correct product or p

roducts of this reaction?

1 answer:

valentinak56 [21]2 years ago

6 0

Answer : The products of this reaction are, $Cu(OH)_2$ and $NaNO_3$

Explanation :

First we have to write the balanced chemical reaction.

$Cu(NO_3)_2+2NaOH\rightarrow Cu(OH)_2+2NaNO_3$

In this reaction, 1 mole of copper nitrate react with the 2 moles of sodium hydroxide to give 1 mole of copper hydroxide and 2 moles of sodium nitrate.

This reaction is a double displacement reaction.

Double displacement reaction : A type of reaction in which a positive cation and a negative anion of two reactants react to exchange their places to form two new products.

The general representation of this reaction is,

$XY+AB\rightarrow XB+AY$

(X and A are the cations, Y and B are the anions)

Hence, the products of this reaction are, copper hydroxide, $Cu(OH)_2$ and sodium nitrate, $NaNO_3$

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A 0.50-kg croquet ball is initially at rest on the grass. When the ball is struck by a mallet, the average force exerted on it i

NeTakaya

The impulse given to the ball is equal to the change in its momentum:

J = ∆p = (0.50 kg) (5.6 m/s - 0) = 2.8 kg•m/s

This is also equal to the product of the average force and the time interval ∆t :

J = F(ave) ∆t

so that if F(ave) = 200 N, then

∆t = J / F(ave) = (2.8 kg•m/s) / (200 N) = 0.014 s

7 0

1 year ago

What is the net charge of a metal ball if there are 21,749 extra electrons in it?

pickupchik [31]

Answer:

$Q=3.47\times 10^{-15}\ C$

Explanation:

Given that,

Number of extra electrons, n = 21749

We need to find the net charge on the metal ball. Let Q is the net charge.

We know that the charge on an electron is $q=1.6\times 10^{-19}\ C$

To find the net charge if there are n number of extra electrons is :

Q = n × q

$Q=21749\times 1.6\times 10^{-19}\ C$

$Q=3.47\times 10^{-15}\ C$

So, the net charge on the metal ball is $3.47\times 10^{-15}\ C$ . Hence, this is the required solution.

6 0

2 years ago

A magnetic field of 37.2 t has been achieved at the mit francis bitter national magnetic laboratory. Find the current needed to

jeka57 [31]

Answer:

Here is the complete question:

https://www.chegg.com/homework-help/questions-and-answers/magnetic-field-372-t-achieved-mit-francis-bitter-national-magnetic-laboratory-find-current-q900632

a) Current for long straight wire $=3.7\ MA$

b) Current at the center of the circular coil $=2.48\times 10^{5}\ A$

c) Current near the center of a solenoid $236.8\ A$

Explanation:

⇒ Magnetic Field due to long straight wire is given by (B),where $B=\frac{\mu\times I}{2\pi(r) },so\ I=\frac{B\ 2\pi(r)}{\mu}$

$\mu=4\pi \times 10^{-7}\ \frac{henry}{m}$

Plugging the values,

Conversion $1\times 10^6 A = 1\ MA$ ,and $2cm=\frac{2}{100}=0.02\ m$

$I=\frac{37.2\times \ 2\pi(0.02)}{4\ \pi \times (10^{-7})}=3.7\ MA$

⇒Magnetic Field at the center due to circular coil (at center) is given by, $B=\frac{\mu\times I (N)}{2(a)}$

So $I= \frac{2B(a)}{\mu\ N} = \frac{2\times 37.2\times 0.42}{4\pi\times 10^{-7}\times 100}=2.48\time 10{^5}\ A$

⇒Magnetic field due to the long solenoid, $B=\mu\ nI=\mu (\frac{N}{l})I$

Then $I=\frac{B}{\mu(\frac{N}{L})} \approx 236.8\A$

So the value of current are $3.7 MA$ , $2.48\times 10^{5} A$ and $236.8\ A$ respectively.

8 0

2 years ago

Help!!!! A loop of area 0.100 m^2 is oriented at 15.5 degree angle to a 0.500 T magnetic field. It rotates until it is at a 45.0

Nat2105 [25]

Answer:

Resulting Change in the Magnetic Flux $=0.013\:Wb$

Explanation:

Thanks!

3 0

2 years ago

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nadezda [96]

Answer:

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4 0

2 years ago