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vfiekz [6]
2 years ago
11

List the symbols for the noble gases

Chemistry
1 answer:
dolphi86 [110]2 years ago
7 0

Answer:here:

Explanation:helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn), and oganesson (Og).

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Disease caused by virus are more dangerous.Why?
nlexa [21]

Answer:

Explanation:

1. Virus's are hard to detect because of their simple construction.

2. Some mutate very easily.

3. It is hard to isolate the virus and kill it without doing damage to the host.

7 0
3 years ago
which of these pieces of equipment would be the most appropriate for precisely measuring 29 mL of liquid? Explain your reasoning
Anna007 [38]

Answer:

The best equipment would be the graduated cylinder. Why?

Firstly, the smallest marking on the graduated cylinder is 2 mL, while on all the others the smallest marking is way above that, like 25 mL and 100 mL.

Without even going into the details, we can first rule out the volumetric flask, since its smallest marking is 100 mL and even that is already bigger than our sample size, hence we would have no markings to accurately measure out 29 mL of our sample had we used the volumetric flask.

Next to be ruled out would be the Erlenmeyer flask, as you can see in the image, it only has three marking, and as the smallest marking is 25 mL, each marking is at least 25 mL, and even so far as going up to 50 mL. This cannot let us accurately measure 29 mL out at all, due to the markings being way too big to do that. Hence, the Erlenmeyer flask is ruled out.

Finally, the beaker seems to be a worthy candidate! Unfortunately, for the same reason as the Erlenmeyer flask, as you can see in the image each marking represents 10 mL. We cannot measure 9 mL in the beaker accurately, and hence the beaker is ruled too.

We are left with the graduated cylinder, and that is our answer.

Explanation:

Hope this helped!

5 0
3 years ago
Apply the given electronegativities to each bond in the examples below to see which are polar and which are nonpolar. Multiple b
Nesterboy [21]

Answer:

H-O-H  polar

O-C-O nonpolar

H-C-N   polar

Explanation:

Looking up the electronegativities of the atoms involved in this question, we have:

Atom   Electronegativity

H               2.2

C               2.55

N               3.04

O               3.44

All of the atoms differ in electronegativity resulting in individual dipole moments in H-O, O-C, H-C and C-N bonds. To find if the molecules will be polar we need to consider the structure of the compound to see if there is a resultant dipole moment.

In H-O-H, we have 2 lone pairs of electrons around the central oxygen atom which push the angle H-O-H  of the ideal tetrahedral structure to be smaller than 109.5 º  resulting in an overall dipole moment making it polar.

In O-C-O, we have two dipole moments that exactly cancel each other in the linear molecule  since the central carbon atom does not have lone pairs of electrons since it  has 2 double bonds. Therefore the molecule is nonpolar.

In H-C-N, again we have have a central carbon atom without lone pairs of electrons and the shape of the molecule is linear. But, now we have that the dipole moment in C-N is stronger than the H-C dipole because of the difference in electronegativity of nitrogen compared to hydrogen. The molecule has an overall dipole moment and it is polar.

8 0
3 years ago
Read 2 more answers
A photon has a frequency of 7.3 × 10–17 Hz. Planck’s constant is 6.63 × 10–34 J•s. The energy of the photon, to the nearest tent
yawa3891 [41]

Given:

E = 7.3 × 10–17 Hz                                                                                      

 h= 6.63 × 10–34 J•s

Now <em>E = hf</em>

where E is the energy of the photon                                                          

h is the Planck's constant                                                                          

f is the frequency of the photon

Substituting the values in the equation we get                                        

E= 7.3 × 10^-17 × 6.63 × 10^-34                                                                  

<u>E= 4.8399 × 10^-50  J. </u>                                                                                                      



7 0
4 years ago
Read 2 more answers
The standard cell potential (E°cell) for the reaction below is +0.63 V. The cell potential for this reaction is __________V when
Elodia [21]

<u>Answer:</u> The cell potential of the above reaction is 0.52 V

<u>Explanation:</u>

The given chemical equation follows:

Pb^{2+}(aq.)+Zn(s)\rightarrow Zn^{2+}(aq.)+Pb(s)

<u>Oxidation half reaction:</u>  Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-

<u>Reduction half reaction:</u>  Pb^{2+}(aq.)+2e^-\rightarrow Pb(s)

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}

where,

E_{cell} = electrode potential of the cell = ? V

E^o_{cell} = standard electrode potential of the cell = 0.63 V

n = number of electrons exchanged = 2

[Zn^{2+}]=1.0M

[Pb^{2+}]=2.0\times 10^{-4}M

Putting values in above equation, we get:

E_{cell}=0.63-\frac{0.059}{2}\times \log(\frac{1.0}{2.0\times 10^{-4}})\\\\E_{cell}=0.52V

Hence, the cell potential of the above reaction is 0.52 V

7 0
3 years ago
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