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azamat
3 years ago
12

Which contains the particles that are vibrating fastest?

Chemistry
2 answers:
Soloha48 [4]3 years ago
6 0
A. The hotter things get the more energy the particles have.
jonny [76]3 years ago
5 0

Answer: Option (a) is the correct answer.

Explanation:

When water is boiled then due to the heat provided vibration between the particles increases. As a result, there will be increase in kinetic energy of particles.

Hence, vibration between the particles of boiling water will be the fastest.

Whereas vibration of particles in frozen water, room temperature water and cold tap water will not be the fastest.

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CAN SOMEONE PLEASEE HELPPP
Masja [62]

Answer:

hope its not to late..............Samira's model correctly demonstrates how the properties changed with the rearrangement of the atoms. However not all atoms are accounted for. There is a missing reactant. Samira's model correctly demonstrated how the atoms in two compounds reacted to form two new products. However, the elements present in the reactants side should be the elements that make up the new products in the product side. But as the diagram shows, Sameera has mistakenly added a new element to one of her products which will be wrong.

Explanation:

8 0
3 years ago
Which ONE of the following is an oxidation–reduction reaction? A) PbCO3(s) + 2 HNO3(aq) ––––> Pb(NO3)2(aq) + CO2(g) + H2O(l)
sveta [45]

Answer:

E) C₂H₄(g) + H₂(g) ⇒ C₂H₆(g)

Explanation:

Which ONE of the following is an oxidation–reduction reaction?

A) PbCO₃(s) + 2 HNO₃(aq) ⇒ Pb(NO₃)₂(aq) + CO₂(g) + H₂O(l). NO. All the elements keep the same oxidation numbers.

B) Na₂O(s) + H₂O(l) ⇒ 2 NaOH(aq). NO. All the elements keep the same oxidation numbers.

C) SO₃(g) + H₂O(l) ⇒ H₂SO₄(aq). NO. All the elements keep the same oxidation numbers.

D) CO₂(g) + H₂O(l) ⇒ H₂CO₃(aq). NO. All the elements keep the same oxidation numbers.

E) C₂H₄(g) + H₂(g) ⇒ C₂H₆(g). YES. <u>C is reduced</u> and <u>H is oxidized</u>.

8 0
3 years ago
A mover packs books, CDs, and DVDs into a moving box. If the box contains 6.5 kg of books, 1.5 kg of CDs, and 2.0 kg of DVDs, wh
mariarad [96]
Given: 
<span>M1 = 6.5 kg of books
</span><span>M2 = 1.5 kg of CDs
</span><span>M3 = 2.0 kg of DVDs

Required: percent by mass of each object

Solution:
First, we calculate the total mass.

M = 6.5 kg + 1.5 kg + 2.0 kg =  10 kg

Percent by mass is calculated by getting the ration of the mass of an object and the total mass multiplied by 100 to get the percent.

%M1 = 6.5 / 10 x 100 = 65%
%M2 = 1.5/10 x 100 = 15%
%M3 = 2.0/10 x 100 = 20%</span>
4 0
3 years ago
Read 2 more answers
I really need help ! This is science
lina2011 [118]

Answer:

7. genetics because hereditary is not the study of hereditary.

8. is hereditary because it is the study of how things pass on to next generation what is inherited.

7 0
3 years ago
Read 2 more answers
For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia
Fittoniya [83]

Answer : The value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction:  Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction:  Cu^{2+}+2e^-\rightarrow Cu

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu

To calculate the E^o_{(Cu^{2+}/Cu)} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}

Putting values in above equation, we get:

1.10V=E^o_{(Cu^{2+}/Cu)}-(-0.76V)

E^o_{(Cu^{2+}/Cu)}=0.34V

Hence, the value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

8 0
3 years ago
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