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Scrat [10]
3 years ago
8

Which method of testing substances is a sure way to identify a chemical reaction?

Physics
2 answers:
marin [14]3 years ago
8 0

Chemical reaction includes chemical changes (modifications in the structures of molecules, new compounds )of the substances in the reaction. As a result of the chemical reaction new substances are formed that cannot be reversed using physical means.

The following method of testing substances is a sure way to identify a chemical reaction:

A. color change

B. Formation of bubbles

D. increase in temperature  

Dmitrij [34]3 years ago
3 0

A change in color hope it helps

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An uncharged capacitor is connected to a 21-V battery until it is fully charged, after which it is disconnected from the battery
attashe74 [19]

Answer:

The voltage bewtween the plates will be 9.5V

Explanation:

Facts:

The capacitance of a parallel plate capacitor having plate area A and plate separation d is C=ϵ0A/d.  

Where ϵ0 is the permittivity of free space.  

A capacitor filled with dielectric slab of dielectric constant K, will have a new capacitance C1=ϵ0kA/d

        C1=K(ϵ0A/d)

        C1=KC   ----------- 1

Where C is the capacitance with no dielectric and C1 is the capacitance with dielectric.

The new capacitance is k times the capacitance of the capacitor without dielectric slab.  

This implies that the charge storing capacity of a capacitor increases k times that of the capacitor without dielectric slab.

Given points

• The terminal voltage of the battery to which the capacitor is connected to charge V=25V

• A dielectric slab of paraffin of dielectric constant K=2  is inserted in the space between the capacitor plates after the fully charged capacitor is disconnected

The charge stored in the original capacitor Q=CV

The charge stored in the original capacitor after inserting dielectric Q1=C1V1

The law of conservation of energy states that the energy stored is constant:

i.e Charge stored in the original capacitor is same as charge stored after the dielectric is inserted.

Q   =  Q1

CV = C1V1

  CV = C1V1  -------2

We derived C1=KC from equation 1. Inserting this into equation 2

   CV = KCV1

 V1 = (CV)/KC

         V/K

       = 21/2.2

      = 9.5

4 0
3 years ago
Calculate the kinetic energy of a 4.00kg object traveling at 16.0 m/s
Elenna [48]
KE = ½mv² = ½(4.00 kg)(16.0 m/s)² = 512 J
3 0
3 years ago
Read 2 more answers
Ablock of mass m2 on arough horinzontal surfaceis connected to aball of mass m1 by alight weight cord over alight weight frictio
aliya0001 [1]
Dndncbsnsndkdkksdkdndnsndnkckd
5 0
3 years ago
How much water will flow in 30 secs through 200 mm of capillary tube of 1.50 mm in diameter, if the pressure difference across t
Paladinen [302]

The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

Qo=1.6 \times 10^{2} \mathrm{~mL}

<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>

\begin{aligned}\Delta P &=6660 \mathrm{~m} / \mathrm{m}^{2} \\\mu &=8.01 \times 10^{-4} \text { Pas } \\t &=30 \mathrm{~s} \\L &=200 \mathrm{~mm}=200 \times 10^{-3} \mathrm{~m} \\D &=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \Rightarrow \gamma=\frac{1.5 \times 10^{-3}}{2} \mathrm{~m}\end{aligned}

Generally, the equation for Rate of flow of Liquid is  mathematically given as

\\$$Q=\frac{\pi r^{4} \times \Delta P}{8 \mu L}

$$

Where dP is pressure difference r is the radius

\mu is the viscosity of water

L is the length of the pipe

Q=\frac{\pi \times\left(\frac{1.5 \times 10^{-3}}{2}\right)^{4} \times 6660}{8 \times 8.01 \times 10^{-4} \times 200 \times 10^{-3}}

Q=5.2 \mathrm{~mL} / \mathrm{s}

In $30s the quantity that flows out of the tube

&Qo=5.2 \times 30 \\&Qo=1.6 \times 10^{2} \mathrm{~mL}

In conclusion, the quantity that flows out of the tube

Qo=1.6 \times 10^{2} \mathrm{~mL}

Read more about the flows rate

brainly.com/question/27880305

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5 0
2 years ago
Jack drops a stone from rest off of the top of a bridge that is 24.4 m above the ground. After the stone falls 6.6 m, Jill throw
alukav5142 [94]
-17.555m/s

first I found the time it took for jacks stone to reach the bottom, using the formula vf = vi + at, vf and vi are final and initial velocities.

then i found the velocity at 6.6m using vf^2 = vi^2 + 2ad
and I found the time it took to get to 6.6m, so that I knew how long Jill waited to throw her stone, I used the formula d = t(vi+vf)/2, then i done total time - the time she waited, to get the time it took for there stones to hit the ground at the same time.

then to find the initial velocity of her throw I used the formula d = vit + (at^2)/2
4 0
3 years ago
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