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Artyom0805 [142]
3 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]3 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]3 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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Estimate the volume of a typical house (2050 feet squared in size and 10 feet tall) answer in units of meters squared
Aloiza [94]

Answer:

Volume = 6248.48 m^{3}

Explanation:

Given:

The area of the house A = 2050\ ft^{2}

The height of the house h=10\ ft

We need to find the volume of a typical house.

Solution:

We find the volume of the house by multiplying the area of the house and height of the house.

Volume = Area\times height

Volume = A\times h

Area and height of the house are known, so we substitute these value in above equation.

Volume = 2050\times 10

Volume = 20500\ ft^{3}

Now we convert the unit from feet to meter.

Divide the volume by 3.2808 for m^{3}

Volume = \frac{20500}{3.2808}

Volume = 6248.48\ m^{3}

Therefore, the volume of the house is 6248.48 m^{3}

8 0
3 years ago
A block with mass m 2.00 kg is placed against a spring on a frictionless incline with angle 30 degrees (Fig. B-43). (The block i
guajiro [1.7K]

Answer:

Explanation:

a )

The stored elastic energy of compressed spring

= 1 / 2 k X²

= .5 x 19.6 x (.20)²

= .392 J

b ) The stored potential energy will be converted into gravitational potential energy of the block earth system when the block will ascend along the incline . So change in the gravitational potential energy will be same as stored elastic potential energy of the spring that is .392 J .

c ) Let h be the distance along the incline which the block ascends.

vertical height attained ( H ) =h sin30

= .5 h

elastic potential energy = gravitational energy

.392 = mg H

.392 = 2 x 9.8 x .5 h

h = .04 m

4 cm .

=

7 0
3 years ago
Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an
ddd [48]

Answer:

Al's mass is 102.92  kg  

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

F_{net} = 0

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

F_{net} = \frac{dp_{horizontal}}{dt} = 0

p_{horizontal_i }= p_{horizontal_f}

where the suffix i and f means initial and final respectively.

The initial momentum will be:

p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}

But, as they are at rest, initially

p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0

p_{horizontal_i} = 0

So, this means:

p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0

We know that the have an combined mass of 195 kg:

m_{total} = m_{Al} + m_{Jo} = 195 \ kg.

so:

m_{Jo} = 195 \ kg - m_{Al}.

m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0

m_{Al} \  v_{Al_f} - m_{Al} \  v_{Jo_f}= - 195 \ kg \  v_{Jo_f}

m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}

m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } {  v_{Al_f} - v_{Jo_f} }

m_{Al} = \frac{195 \ kg  \ v_{Jo_f} } {    v_{Jo_f} - v_{Al_f} }

Now, we can use the values:

v_{Al_f}= 10.2 \frac{m}{s}

v_{Jo_f}= - 11.4 \frac{m}{s}

where the minus sign appears as they are moving at opposite directions

m_{Al} = \frac{195 \ kg  ( - 11.4 \frac{m}{s} ) } {   (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }

m_{Al} = 102.92 \ kg

and this is the Al's mass.

5 0
3 years ago
Un cuerpo se encuentra en reposo sobre una mesa horizontal. Entonces se puede afirmar que:
vazorg [7]

Answer:

C) solo III

Explanation:

Para solucionar este problema debemos analizar cada una de las opciones hasta llegar a la opcion valida.

I) el cuerpo pesa igual que su masa.

Esta opcion no puede ser ya que el peso de un cuerpo se define como el producto de la masa por la aceleracion gravitacion.

w=m*g

donde:

w = peso [N]

m = masa [kg]

g = aceleracion gravitacional = 9.81 [m/s²]

Como podemos ver el peso siempre sera mayar que la masa, ya que el peso es resultado de la multiplicacion de la masa por la gravedad.

II) Por medio de un analisis de fuerzas en el eje-y, la fuerza del peso se dirige hacia abajo mientras que la fuerza normal tiene igual magnitud, pero se dirige hacia arriba. Por esto la segunda opcion no puede ser.

III) El cuerpo se encuentra en equilibrio, es decir las unicas fuerzas que actuan sobre el cuerpo son el peso y la fuerza normal. Pero estas fuerzas son iguales y opuestas en direccion, por la tanto se cancelan y estan en equilibrio.

Esta es la opcion valida, la fuerza neta es nula.

5 0
3 years ago
Señalar la importancia de las capacidades fisico-motiz que se desarrollan en el futbol de salon y dar un ejemplo para cada uno
vekshin1

La respuesta correcta para esta pregunta abierta es la siguiente.

A pesar de que no anexas opciones o incisos para responder, podemos comentar lo siguiente.

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Estas capacidades físico-motrices son las que le permiten a un jugador realizar su máximo esfuerzo, mejorar su desempeño físico y conseguir resultados positivos.

Estamos hablando de la fuerza, la velocidad y la resistencia.

La velocidad es la aceleración que el jugador de futbol necesita para aumentar su velocidad de un punto A, a un punto B, en el menor tiempo posible.

La resistencia es la capacidad del jugador de futbol para mantener ese nivel de aceleración y desempeño, sin bajar su rendimiento. Su capacidad física debe ser resistente para ser constante en su rendimiento físico.

La fuerza es la potencia con la que desempeña los movimiento físicos dentro de la cancha.

7 0
3 years ago
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