Which of the following is an example of a chemical change?

Find the height of the coconut from the ground at the moment the arrow hit it.

Wewaii [24]

Answer:

$v = u + at \\ 0 = 45 \sin(20) - 9.81 \times t \\ t = 1.56s$

7 0

2 years ago

Calculate the total number of Cl atoms in 150mL of liquid Ccl4 (d=1.589g/mL)

GalinKa [24]

Answer:

The total number of Cl atoms in 150mL of liquid CCl4 is 3.73*10²⁴.

Explanation:

First you must determine the mass of CCL4 present in 150mL of CCl4. Density is a quantity that allows us to measure the amount of mass in a certain volume of a substance, whose expression for its calculation is the quotient between the mass of a body and the volume it occupies:

$density=\frac{mass}{volume}$

In this case, the density value of d = 1.589 g/mL. Then, being the volume equal to 150 mL, the value of the mass can be calculated as:

mass= density*volume

mass=1.589 g/mL * 150 mL

mass= 238.35 g

Now, being the molar mass of CCl4 154 g/mol, the number of moles that 238.35 g represents is calculated as:

$moles=\frac{238.35 g}{154 \frac{g}{mol} }$

moles= 1.55

1 mole of the compound CCl4 contains 4 moles of Cl. Then, using a simple rule of three, it is possible to calculate the number of moles of Cl that 1.55 moles of CCl4 contain:

$moles of Cl=\frac{1.55 moles of CCl_{4} *4 moles of Cl}{1 mole of CCl_{4} }$

moles of Cl= 6.2

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023*10²³ particles per mole. Avogadro's number applies to any substance. In this case it can be applied as follows: if 1 mole of Cl contains 6.023*10²³ atoms, 6.2 moles of Cl how many atoms does it contain?

$atoms of Cl=\frac{6.2 moles*6.023*10^{23} atoms}{1 mole}$

atoms of Cl= 3.73*10²⁴

The total number of Cl atoms in 150mL of liquid CCl4 is 3.73*10²⁴.

8 0

3 years ago

4.-Una atleta patea un balon (masa de 650gr y angulo de 45 grados) arriba de un edificio con una velcidad 21 m/s, el edificio mi

marta [7]

Answer:

v = 25 m/s

Explanation:

½mv₀² + mgh = ½mv₁²

v₁² = v₀² + 2gh

v₁ = √(21² + 2(9.8)(10))

v₁ = 25.238858..

3 0

3 years ago

What could be the possible answer to the question ? thankyou ~

Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

The value of F₀ so that string 1 remains vertical is approximately 0.377·M·g

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

$T_1 = \mathbf{\dfrac{M \cdot g}{2}}$

Which gives;

$M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}$

$T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})} = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}$

F₀ = T₂·sin(37°)

Which gives;

$F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2} \approx \mathbf{0.377 \cdot M \cdot g}$