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3 years ago

11

From the options provided for each element below, choose the properties that it may have based on its location in the periodic t

able. Neon (Ne):
A,) shiny

B.)a conductor

C.)a gas

ANSWER IS: option C

2 answers:

Zarrin [17]3 years ago

8 0

It is answer letter c

liq [111]3 years ago

6 0

The answer is letter c

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Setting up a one-step unit conversion A student sets up the following equation to convert a measurement. (The ? stands for a num

Aneli [31]

Answer:

0.0008 m

Explanation:

We are given that 0.080 cm

We have to convert 0.080 cm into meter

To find the value of 0.080 cm in meter we are using unitary method

We know that

100 cm =1 m

1 cm = $\frac{1}{100} m$

Therefore, 0.080 cm = $\frac{1}{100}\times 0.080 m$

0.080 cm = $\frac{1}{100}\times \frac{80}{1000}m$

0.080 cm = $\frac{8}{10000}m$

0.08cm=0.0008 m

Hence, the value of 0.080 cm is equal to 0.0008 meter .

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3 years ago

aev [14]

Answer:

Copper(II) sulphate – sodium hydroxide reaction

The reaction between copper(Il) sulphate and sodium hydroxide solutions is a good place to start. If you slowly add one to the other while stirring, you will get a precipitate of copper(II) hydroxide, Cu(OH)2.

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3 years ago

PPPPPPPPLLLLLLLLLLLLLLZZZZZZZZZZ HELP ME ITS ALREDY OVER DUE PLZ HELP ME PPPLLLZZZ I NEED IT IM DESPRET

nikitadnepr [17]

Answer:

1) Big Bang Theory

2) we are in a phase of the big bang theory

Explanation:

So for number 1 the answer is the -big bang theory- and number 2 there is a theory that pretty much says that were in a constant cycle. what I mean is theoretically in a couple million years the entire universe will form together again into another dense ball and repeat the process. So basically the universe is still in the "expanding phase" and that's what backs up the theory of the big bang (the fact that were in one of the big bang theory's "phases")

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2 years ago

Read 2 more answers

if 100. mL of 0.800 M Na2SO4 is added to 200. mL of 1.20 M NaCl, what is the concentration of Na+ ions in the final solution? As

Black_prince [1.1K]

Compounds Na₂SO₄ and NaCl are mixed together are we are asked to find the concentration of Na⁺ in the mixture
Na₂SO₄ ---> 2 Na⁺ + SO₄³⁻

1 mol of Na₂SO₄ gives out 2 mol of Na⁺ ions

the number of Na₂SO₄ moles added - 0.800 M/1000 * 100 ml
= 0.08 mol
therefore number of Na⁺ ions from Na₂SO₄ = 0.08 * 2 = 0.16 mol

NaCl ----> Na⁺ + Cl⁻
1 mol of NaCl gives 1 mol of Na⁺ ions
number of NaCl moles added = 1.20 M/1000 * 200 ml
= 0.24 mol
number of Na⁺ ions from NaCl = 0.24 mol

total number of Na⁺ ions in the mixture = 0.16 mol + 0.24 mol = 0.4 mol
as stated the volumes are additive,
therefore total volume = 100 ml + 200 ml = 300 ml
the concentration of Na⁺ ions = number of moles / volume
= 0.4 mol/ 0.3 dm³
concentration of Na⁺ = 1.33 mol/dm³

5 0

3 years ago

An atom has a diameter of 2.50 Å and the nucleus of that atom has a diameter of 9.00×10−5 Å . Determine the fraction of the volu

chubhunter [2.5K]

Answer:

The fraction of the volume of the atom that is taken up by the nucleus is $4.6656\times 10^{-14}$ .

The density of a proton is $6.278\times 10^{14} g/cm^3$ .

Explanation:

Diameter of the atom ,d = 2.50 Å

Radius of the atom ,r = 0.5 d=0.5 × 2.50 Å = 1.25Å

Volume of the sphere= $\frac{4}{3}\pi r^3$

Volume of atom = V

$V=\frac{4}{3}\pi r^3$ ..[1]

Diameter of the nucleus ,d' = $9.00\times 10^{-5}\AA$

Radius of the nucleus ,r' = 0.5 d'= $0.5\times 9.00\times 10^{-5}\AA=4.5\times 10^{-5}\AA$

Volume of nucleus = V'

$V=\frac{4}{3}\pi r'^3$ ..[2]

Dividing [2] by [1]

$\frac{V'}{V}=\frac{\frac{4}{3}\pi r'^3}{\frac{4}{3}\pi r^3}$

$=\frac{r'^3}{r^3}=\frac{(4.5\times 10^{-5}\AA)^3}{(1.25 \AA)^3}$

$\frac{V'}{V}=4.6656\times 10^{-14}$

The fraction of the volume of the atom that is taken up by the nucleus is $4.6656\times 10^{-14}$ .

Diameter of the proton ,d = $1.72\times 10^{-15} m = 1.72\times 10^{-13} cm$

1 m = 100 cm

Radius of the proton,r = 0.5 d= $0.5\times 1.72\times 10^{-13} cm=8.6\times 10^{-14} cm$

Volume of the sphere= $\frac{4}{3}\pi r^3$

Volume of atom = V

$V=\frac{4}{3}\times 3.14\times (8.6\times 10^{-14} cm)^3=2.664\times 10^{-39}cm^3$

Mass of proton, m = 1.0073 amu = $1.0073\times 1.66054\times 10^{-24} g$

$1 amu = 1.66054\times 10^{-24} g$

Density of the proton : d

$d=\frac{m}{V}=\frac{1.0073\times 1.66054\times 10^{-24} g}{2.664\times 10^{-39}cm^3}=6.278\times 10^{14} g/cm^3$

The density of a proton is $6.278\times 10^{14} g/cm^3$ .

5 0

3 years ago