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Artyom0805 [142]
4 years ago
13

Identify the reaction type for each generic chemical equation. A + B → AB: AB → A + B: Hydrocarbon + O2 → CO2 + H2O: AB + CD →

AD + CB:
Physics
2 answers:
Ghella [55]4 years ago
6 0

Answer:

1. A+B\rightarrow AB: Synthesis

AB\rightarrow A+B: Decomposition

Hydrocarbon+O_2\rightarrow CO_2+H_2O:  Combustion

4. AB+CD\rightarrow AD+CB : Double displacement

Explanation:

1. Synthesis reaction is a chemical reaction in which two reactants are combining to form one product.

Example: Li_2O+CO_2\rightarrow Li_2CO_3  

2. Decomposition is a type of chemical reaction in which one reactant gives two or more than two products.

Example: Li_2CO_3\rightarrow Li_2O+CO_2

3. Combustion is a type of chemical reaction in which hydrocarbon is reacted with oxygen to form carbon dioxide and water.

Example:  

CH_4+2O_2 \rightarrow CO_2+2H_2O

4. Double displacement reaction is one in which exchange of ions take place.  

Example: NaOH+HCl\rightarrow NaCl+NH_2O

777dan777 [17]4 years ago
6 0

Answer:

Identify the reaction type for each generic chemical equation.

A + B → AB:  synthesis

AB → A + B:  decomposition

Hydrocarbon + O2 → CO2 + H2O:  combustion

AB + CD → AD + CB:   replacement

Explanation:

idk

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Two power lines run parallel for a distance of 222 m and are separated by a distance of 40.0 cm. if the current in each of the t
earnstyle [38]
1) Magnitude of the force:

The magnetic field generated by a current-carrying wire is
B= \frac{\mu_0I}{2 \pi r}
where
\mu_0 is the vacuum permeability
I is the current in the wire
r is the distance at which the field is calculated

Using I=135 A, the current flowing in each wire, we can calculate the magnetic field generated by each wire at distance 
r=40.0 cm=0.40 m, 
which is the distance at which the other wire is located:
B= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} N/A^2)(135 A) }{2 \pi (0.40 m)}=6.75 \cdot 10^{-5} T

Then we can calculate the magnitude of the force exerted on each wire by this magnetic field, which is given by:
F=ILB=(135 A)(222 m)(6.75 \cdot 10^{-5}T)=2.03 N

2) direction of the force: 
The two currents run in opposite direction: this means that the force between them is repulsive. This can be determined by using the right hand rule. Let's apply it to one of the two wires, assuming they are in the horizontal plane, and assuming that the current in the wire on the right is directed northwards:
- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)

Now let's apply the right-hand rule to the wire on the right:
- index finger: current --> northward
- middle finger: magnetic field --> upward
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A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.
6 0
3 years ago
The electric field in a region is uniform (constant in space) and given by E-( 148.0 1 -110.03)N/C. An additional charge 10.4 nC
enyata [817]

Answer:

The y-component of the electric force on this charge is F_y = -1.144\times 10^{-6}\ N.

Explanation:

<u>Given:</u>

  • Electric field in the region, \vec E = (148.0\ \hat i-110.0\ \hat j)\ N/C.
  • Charge placed into the region, q = 10.4\ nC = 10.4\times 10^{-9}\ C.

where, \hat i,\ \hat j are the unit vectors along the positive x and y axes respectively.

The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,

\vec E = \dfrac{\vec F}{q}\\\therefore \vec F = q\vec E\\=(10.4\times 10^{-9})\times (148.0\ \hat i-110.0\ \hat j)\\=(1.539\times 10^{-6}\ \hat i-1.144\times 10^{-6}\ \hat j)\ N.

Thus, the y-component of the electric force on this charge is F_y = -1.144\times 10^{-6}\ N.

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