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Semmy [17]
2 years ago
13

The country like Nepal with diversity must have competition and cooperation justify​

Chemistry
1 answer:
Len [333]2 years ago
3 0
Help????? You need to write as if you are Dmitri Mendeleev and you have just made your proposed version of the periodic table. You need to write to the Royal Society of Chemistry in London explaining:
• Who you are
• What your periodic table is like, the groups, elements and features
• Why you think your periodic table is correct
• How you have built on the work of others or why you think their work is not
correct
You might be interested in
PLEASE HELP 20 points
777dan777 [17]

  n = 1.5atm (15L) / .0821 (280k) = .98 mol NaCl

  NaCl = 22.99g Na + 35.45g Cl = 58.44g NaCl

  58.44g NaCl x .98 mol NaCl = 57.27g NaCl

Explanation:

hope you get it right :)

4 0
2 years ago
The specific heat of aluminum is 0.214 cal/g.oC. Determine the energy, in calories, necessary to raise the temperature of a 55.5
Natasha2012 [34]
For this problem, we use the formula for sensible heat which is written below:

Q= mCpΔT
where Q is the energy
Cp is the specific heat capacity
ΔT is the temperature difference

Q = (55.5 g)(<span>0.214 cal/g</span>·°C)(48.6°C- 23°C)
<em>Q = 304.05 cal</em>
4 0
3 years ago
3-chloropropylamine what is?
marta [7]

Answer:

a chemical compound

6 0
1 year ago
2. Calculate the atomic mass of an element that has two isotopes, each with 50.00% abundance. One isotope has a mass of 63.00 am
melamori03 [73]

Answer:

The atomic mass of element is 65.5 amu.

Explanation:

Given data:

Abundance of X-63 = 50.000%

Atomic mass of  X-63 = 63.00 amu

Atomic mass of X-68 = 68.00 amu

Atomic mass of element = ?

Solution:

Abundance of X-68 = 100-50 = 50%

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

Average atomic mass  = (50×63)+(50×68) /100

Average atomic mass =  3150 + 3400 / 100

Average atomic mass  = 6550 / 100

Average atomic mass = 65.5 amu.

The atomic mass of element is 65.5 amu.

7 0
3 years ago
When would formation of a derivative be a necessary step in this experiment?
const2013 [10]
<span>The formation of a derivative being a necessary step in the experiment lies in the importance of the derived structure. Often the derived product confers to reaction pathways which uses less reactive starting materials and more easily proceeds to completion. This also allows us to take a small amount of sample. The derived product at times is a general compound allowing its easy analysis. Often we encounter a product but we find it difficult to analyse it in ways we want. Here lies the essence of forming a derivative which often are simpler compounds allowing easier analysis yet having similar functional groups and structural properties. Also sometimes we encounter problems when our desired product is unstable and forms stable degraded products. But if we somehow manage to synthesize a derivative it may be relatively stable and form no degradation products. It would be stable at least for a significant period of time making it easier to study its properties. The derived product also at times are synthesized using general reaction pathways facilitating a way of easier synthesis and helping it to correlate with other similar reaction pathways and products.So the above paragraph accounts for the need of derivatives. When we encounter problems similar to those mentioned above it becomes necessary for a researcher to form rather synthesize a derivative.</span>
6 0
3 years ago
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