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Tasya [4]
3 years ago
11

Helium is a colorless gas that does not react with other elements. Using what you know about valence electrons, which other elem

ents would you expect to have similar properties as Helium? Explain.
Chemistry
1 answer:
serg [7]3 years ago
8 0

Answer:

Neon, Argon, Krypton, Xenon and Radon

Explanation:

The other elements that would be expected to have similar properties as the Helium from the periodic table are the noble gases.

The noble gases are a special group of elements belonging to the last group of the periodic table.

They are called the O group or inert group because elements in this group do not combine with other atoms due to their completely filled outermost shell.

This makes them unreactive and do not readily combine with other atoms.

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Do all colors of light travel at the same speed
bogdanovich [222]

Answer:

No because some lights we cannot see because they move to slow or they move to fast for our eye to see. There is just a small little gap compared to what light rays we can actually see. In this picture you can see what I mean.

Explanation:

5 0
4 years ago
Read 2 more answers
The number at the end of an isotope’s name is the ____ number. A) Atomic B)neutron C)mass D)electron
storchak [24]
 each isotope of a given element has a different mass number. For example, carbon-12, carbon-13<span> and carbon-</span>14<span> are </span>three<span> isotopes of the element carbon with mass numbers </span>12<span>, </span>13<span> and </span>14<span> </span>
3 0
3 years ago
Read 2 more answers
Be sure to answer all parts. Calculate the molality, molarity, and mole fraction of FeCl3 in a 24.0 mass % aqueous solution (d =
Anna71 [15]

Answer:

m= 1.84 m

M= 1.79 M

mole fraction (X) =

Xsolute= 0.032

Xsolvent = 0.967

Explanation:

1. Find the grams of FeCl3 in the solution: when we have a mass % we assume that there is 100 g of solution so 24% means 24 g of FeCl3 in the solution. The rest 76 g are water.

2. For molality we have the formula m= moles of solute / Kg solvent

so first we pass the grams of FeCl3 to moles of FeCl3:

24 g of FeCl3x(1 mol FeCl3/162.2 g FeCl3) = 0.14 moles FeCl3

If we had 76 g of water we convert it to Kg:

76 g water x(1 Kg of water/1000 g of water) = 0.076 Kg of water

now we divide m = 0.14 moles FeCl3/0.076 Kg of water

m= 1.84 m

3. For molarity we have the formula M= moles of solute /L of solution

the moles we already have 0.14 moles FeCl3

the (L) of solution we need to use the density of the solution to find the volume value. For this purpose we have: 100 g of solution and the density d= 1.280 g/mL

The density formula is d = (m) mass/(V) volume if we clear the unknown value that is the volume we have that (V) volume = m/d

so V = 100 g / 1.280 g/mL = 78.12 mL = 0.078 L

We replace the values in the M formula

M= 0.14 moles of FeCl3/0.078 L

M= 1.79 M

3. Finally the mole fraction (x)  has the formula

X(solute) = moles of solute /moles of solution

X(solvent) moles of solvent /moles of solution

X(solute) + X(solvent) = 1

we need to find the moles of the solvent and we add the moles of the solute like this we have the moles of the solution:

76 g of water x(1 mol of water /18 g of water) = 4.2 moles of water

moles of solution = 0.14 moles of FeCl3 + 4.2 moles of water = 4.34 moles of solution

X(solute) = 0.14 moles of FeCl3/4.34 moles of solution = 0.032

1 - X(solute) = 1 - 0.032 = 0.967

6 0
3 years ago
The chemical bonds of carbohydrates and lipids have high potential energy because:
gladu [14]

Answer:

c. Many of their bonds are C-C and C-H

Explanation:

The majority of bonds in  carbohydrates and lipids( being an organic compound) are C-C and C-H. Like glucose, fructose or galactose ,etc.

These bonds are strong and do require a lot of energy to break. Thus, a lot of energy are required to break carbs and lipids into simpler compounds.Therefore, carbohydrates and lipids have high potential energy.

The correct answer is c.

4 0
3 years ago
An unknown compound contains 38.7 % calcium, 19.9 % phosphorus and 41.2 % oxygen. what is the empirical formula of this compound
Temka [501]

The amount per 100 g is:

38.7 % calcium = 38.7g Ca / 100g compound = 38.7g

19.9 % phosphorus = 19.9g P / 100g compound = 19.9g

41.2 % oxygen = 41.2g O / 100g compound = 41.2g

The molar amounts of calcium, phosphorus and oxygen in 100g sample are calculated by dividing each element’s mass by its molar mass:

Ca = 38.7/40.078 = 0.96

P = 19.9/30.97 = 0.64

O = 41.2/15.99 = 2.57

C0efficients for the tentative empirical formula are derived by dividing each molar amount by the lesser value that is 0.64 and in this case, after that multiply wih 2.

Ca = 0.96 / 0.64 = 1.5=1.5 x 2 = 3

P = 0.64 / 0.64 = 1 = 1x2= 2

O = 2.57 / 0.64 = 4= 4x2= 8

Since, the resulting ratio is calcium 3, phosphorus 2 and oxygen 8

<span>So, the empirical formula of the compound is Ca</span>₃(PO₄)₂

5 0
4 years ago
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