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3 years ago

Changing subscripts is one of the steps of balancing a chemical equation

Physics

1 answer:

Sergeeva-Olga [200]3 years ago

7 0

Answer:

No, Changing subscripts is not the step of balancing a chemical equation

Explanation:

In balancing a chemical equation you have to keep one thing in mind that the no of atoms and elements on both sides such as reactant (before reaction) and product( after a reaction) must be equal. Coefficients are added to equal the no of atoms and elements, but we will not change the subscript.

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A manometer that is filled with water (rho=1000 kg/m3) is connected to a wind tunnel to measure the pressure inside the test sec

Anna [14]

To develop this problem it is necessary to apply the concepts performed to the absolute pressure based on the reference pressure (atmospheric) and the pressure that is generated due to the height of the column of the measured liquid.

In mathematical terms the previous concept can be expressed as

$P_{abs} = P_{atm} -\rho gh$

Where

$P_{atm}=$ Atmospheric Pressure

$\rho =$ Density

g = Gravitational acceleration

h = Height

Our values are given as

$h = 10*10^{-3}m$

g = 9.8m/s

$\rho_{water} = 1000kg/m^3$

$P_{atm} = 100kPa$

Replacing we have then that

$P_{abs} = P_{atm} -\rho gh$

$P_{abs} = 100 -(1000)(9.8)(10*10^{-3})$

$P_{abs} = 99.9019kPa$

Therefore the absolute pressure in the test section is 99.9019kPa

6 0

3 years ago

a force 2.4E2 N exists between a positive charge of 8E-5 C and a positive charge of 3E-5 C. What distance separates the charges?

Verdich [7]

The distance between the two charges is $0.3 m$

Explanation:

The electrostatic force between two charged objects is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the charges of the two objects

r is the separation between the two charges

In this problem, we are given the following:

$q_1 = 8\cdot 10^{-5} C$

$q_2 = 3\cdot 10^{-5} C$

$F=2.4\cdot 10^2 N$

Therefore, we can rearrange the equation to solve for r, the distance between the two charges:

$r=\sqrt{\frac{kq_1 q_2}{F}}=\sqrt{\frac{(8.99\cdot 10^9)(8\cdot 10^{-5})(3\cdot 10^{-5})}{2.4\cdot 10^2}}=0.3 m$

Learn more about electrostatic force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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3 years ago

A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot was in view for a total of 0.51 s,

Gelneren [198K]

The pot spends T = 0.185s going up and 0.185s going down past the window.
The average speed passing by the window is 2.20 m/0.185s = 11.89 m/s.
During passage, the pot increases speed by T*g = 0.185*9.81 = 1.815 m/s
The speed is therefore 12.80 m/s at the bottom of the window and 10.98 m/s at the top of the window.

The 10.98 m/s speed at the top of the window allows it to rise another 10.98^2/(2g)= 6.15 m past the top of the window

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3 years ago

A 0.500-kg mass suspended from a spring oscillates with a period of 1.18 s. How much mass must be added to the object to change

olga55 [171]

Answer:

The add mass = 5.465 kg

Explanation:

Note: Since the spring is the same, the length and Tension are constant.

f ∝ √(1/m)........................ Equation 1 (length and Tension are constant.)

Where f = frequency, m = mass of the spring.

But f = 1/T ..................... Equation 2

Substituting Equation 2 into equation 1.

1/T ∝ √(1/m)

Therefore,

T ∝ √(m)

Therefore,

T₁/√m₁ = k

where k = Constant of proportionality.

T₁/√m₁ = T₂/√m₂ ........................ Equation 3

making m₂ the subject of the equation

m₂ = T₂²(m₁)/T₁²........................... Equation 4

Where T₁ = initial, m₁ = initial mass, T₂ = final period, m₂ = final mass.

Given: T₁ = 1.18 s m₁ = 0.50 kg, T₂ = 2.07 s.

Substituting into equation 4

m₂ = (2.07)²(0.5)/(1.18)²

m₂ = 4.285(1.392)

m₂ = 5.965 kg.

Added mass = m₂ - m₁

Added mass = 5.965 - 0.5

Added mass = 5.465 kg.

Thus the add mass = 5.465 kg

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3 years ago

A 3-kg chunk of putty moving at 2 m/s collides with and sticks to a 5-kg bowling ball initially at rest. The bowling ball and pu

Oduvanchick [21]

Answer:

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4 0

4 years ago