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julia-pushkina [17]
3 years ago
15

Please Please Please Please Please Help Me (please answer this question completely and 1 will friend you and help you with your

question.)
How do static charges interact, and how are they released?
Use the sentence starters provided to help you answer the scientific explanation.

Static charges interact by ___________________________.

Static charges are released when ___________________________.

My scientific explanation is:
Your answer should include at least 6 complete sentences to explain the claim, evidence and reasoning.
Be sure to check your grammar and spelling.
Chemistry
1 answer:
pav-90 [236]3 years ago
6 0

Answer:

Static electricity is a type of energy that is produced by friction.

Explanation:

Static electricity may be the result when there is an imbalance between the negative charges and the positive charges in the surroundings. They are non contact forces. They pull of push without actually touching the body. The charged particles interact by pulling or pushing the uncharged particles.

These charges are released when they come closer to some uncharged or other charged particles like the electrostatic charges. It is the charge build up in an area.

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For the reaction A +B+ C D E, the initial reaction rate was measured for various initial concentrations of reactants. The follow
lora16 [44]

Answer:

Rate constant of the reaction is 3.3\times 10^{-3} M^{-2} s^{-1}.

Explanation:

A + B + C → D + E

Let the balanced reaction be ;

aA + bB + cC → dD + eE

Expression of rate law of the reaction will be written as:

R=k[A]^a[B]^b[C]^c

Rate(R) of the reaction in trail 1 ,when :

[A]=0.30 M,[B]=0.30 M,[C]=0.30 M

R=9.0\times 10^{-5} M/s

9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c...[1]

Rate(R) of the reaction in trail 2 ,when :

[A]=0.30 M,[B]=0.30 M,[C]=0.90 M

R=2.7\times 10^{-4} M/s

2.7\times 10^{-4} M/s=k[0.30 M]^a[0.30 M]^b[0.90 M]^c...[2]

Rate(R) of the reaction in trail 3 ,when :

[A]=0.60 M,[B]=0.30 M,[C]=0.30 M

R=3.6\times 10^{-4} M/s

3.6\times 10^{-4} M/s=k[0.60 M]^a[0.30 M]^b[0.30 M]^c...[3]

Rate(R) of the reaction in trail 4 ,when :

[A]=0.60 M,[B]=0.60 M,[C]=0.30 M

R=3.6\times 10^{-4} M/s

3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c...[4]

By [1] ÷ [2], we get value of c ;

c = 1

By [3] ÷ [4], we get value of b ;

b = 0

By [2] ÷ [3], we get value of a ;

a = 2

Rate law of reaction is :

R=k[A]^2[B]^0[C]^1

Rate constant of the reaction = k

9.0\times 10^{-5} M/s=k[0.30 M]^2[0.30 M]^0[0.30 M]^1

k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}

k=3.3\times 10^{-3} M^{-2} s^{-1}

7 0
3 years ago
The molar madd of Ca(OH)2
Nonamiya [84]
What you have to do is find a periodic table and add the mass of each atom that the compound is made of.

Ca= 40.1
O= 16.0
H= 1.01

keep in mind that you have to also account for how many atoms of each there are in the molecule. for example, in Ca(OH)2, there are one Ca, two O and two H

so the molar mass of Ca(OH)2= 40.1 + (2 x 16.0) + (2 x 1.01)= 74.12 g/mol
8 0
3 years ago
3. What is the name of the technique used to weigh the unknown liquid in part A of the
Anna [14]

Answer:

The technique is called weighing by difference

7 0
3 years ago
Which provides a baseline for an experimental investigation
Goryan [66]

Answer:

control group

Explanation:

7 0
2 years ago
Which of the following statements concerning equilibrium constants is true? The equilibrium constant of the reverse reaction is
Brut [27]

Answer:

When the coefficients in a balanced chemical reaction are multiplied by two, the equilibrium constant is not affected.

Explanation:

The equilibrium constant of a reaction is known to remain steady.

Even if all the coefficients of all the species in the reaction are multiplied by two, the value of the equilibrium constant will reamin the same because the equilibrium position will not change as a result of that.

3 0
3 years ago
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