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iogann1982 [59]
3 years ago
10

Which of the following is used in pencils?

Chemistry
2 answers:
goldenfox [79]3 years ago
8 0

Answer:

the first is graphite........

Mamont248 [21]3 years ago
3 0

Answer:

1st answer:  A. Graphite

2nd answer:  B.  Mercury

Explanation:

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I have some questions, any help would be loved!
Bingel [31]

Answer: vv

Explanation:

8 0
2 years ago
For group 1a metals, which electron is the most difficult to remove?
shtirl [24]

The periodic table contains groups and periods that include the elements. For group 1 metal lithium is least likely to lose an electron.

<h3>What are group 1 metals?</h3>

Group 1 metals are the alkali metals that include, Li, Na, K, Rb, and Cs. They show the property exhibited by the metals. The chemical trends of group 1 show that cesium loses an electron more easily than the other elements.

When going down the group the tendency to lose electrons increases as the atomic radius increases. The electron gets far away from the nucleus making it easy to get removed.

Therefore, lithium being the first element of the group has the smallest radii and is least likely to lose an electron.

Learn more about the group I metals here:

brainly.com/question/27187436

#SPJ4

4 0
2 years ago
You decided to prepare a phosphate buffer from solid sodium dihydrogen phosphate (NaH2PO4) and disodium hydrogen phosphate (Na2H
KIM [24]

Answer:

For disodium hydrogen phosphate:

5.32g Na2HPO4

For sodium dihydrogen phosphate:

7.65g Na2HPO4

Explanation:

First, you have to put all the data from the problem that you going to use:

-NaH2PO4 (weak acid)

-Na2HPO4 (a weak base)

-Volume = 1L

-Buffer pH = 7.00

-Concentration of [NaH2PO4 + Na2HPO4] = 0.100 M

What we need to find the pKa of the weak acid, in this case NaH2PO4, for that you need to find the Ka (acid constant) of NaH2PO4, and for this we use the pKa of the phosphoric acid as follow:

H3PO4 = H2PO4 + H+    pKa1 = 2.14

H2PO4 = HPO4 + H+       pKa2 = 6.86

HPO4 = PO4 + H+      pKa3 = 12.4

So, for the preparation of buffer, you need to use the pKa that is near to the value of the pH that you want, so the choice will be:

pKa2= 6.86

Now we going to use the Henderson Hasselbalch equation for the pH of a buffer solution:

pH = pKa2 + log [(NaH2PO4)/(Na2HPO4)]

The solution of the problem is attached to this answer.

Download odt
7 0
3 years ago
A ray of yellow light has a wavelength of about 5.8×10−7 m. Will exposure to yellow light cause electrons to be emitted from ces
Dima020 [189]
Given the wavelength of the yellow light (700 nm. in this case) we can find the frequency 
<span>by dividing the speed of light c by the wavelength w, that is: f = c/w and we know that </span>
<span>c is equal to 2.998 * 10**8 meters per second. </span>
<span>So the frequency f = (2.998 * 10**8) / (7.0 * 10**-7) = 4.283 * 10**14 cycles per sec. </span>
<span>(or Hz.) Since the threshold frequency of Cs is 9.39 * 10**14 Hz, the red light doesn't </span>
<span>have a high enough frequency (or energy) to cause electron emission. </span>
<span>Hope this answers your question.</span>
8 0
3 years ago
Will give brainliest!!!!
Alekssandra [29.7K]

Answer:

Its B :)

Hope this helps

7 0
3 years ago
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