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3 years ago

What is the physical state of helium at room temperature

1 answer:

3 0

Depending on the exact room temp, say the helium is in a plastic (one of the ones you buy already full of helium) balloon. Now, you should notice the balloon may be a little squishy, but it is still afloat and looking happy. If you took it outside on a cold day, it would deflate, but when you take it back inside where it is warm, it will reinflate. This is called Boyle's Law. Think if it as if the Helium molecules are partying in the balloon, but when they go out in the cold, they huddle together for warmth.

You might be interested in

If you dissolve 3.5 moles of cacl2 in solution how many moles of ca2+ ions and cl- ions will there be in the solution

Wittaler [7]

Balanced chemical reaction (dissociation) : CaCl₂(aq) → Ca²⁺(aq) + 2Cl⁻(aq).
n(CaCl₂) = 3,5 mol.
From chemical reaction: n(CaCl₂) : n(Ca²⁺) = 1 : 1.
n(Ca²⁺) = 3,5 mol.
From chemical reaction: n(CaCl₂) : n(Cl⁻) = 1 : 2.
n(Cl⁻) = 2 · 3,5 mol.
n(Cl⁻) = 7 mol.
Calcium chloride is salt that completely dissociate in water on calcium cations and chlorine anions.

8 0

4 years ago

You perform a combustion analysis on a 255 mg sample of a substance that contains only C, H, and O, and you find that 561 mg of

zhuklara [117]

The combustion of an organic compound is mostly written as,
CaHbOc + O2 --> CO2 + H2O
where a, b, and c are supposed to be the subscripts of the elements C, H, and O in the compound. Determining the number of moles of C and H in the product which is the same as that in the compound,
(Carbon, C) : (561 mg) x (12/44) = 153 mg x (1 mmole/12 mg) = 12.75
(Hydrogen, H) : (306 mg) x (2/18) = 34 mg x (1 mmole/1 mg) = 34
Calculating for amount of O in the sample,
(oxygen, O) = 255 - 153 mg - 34 mg = 68 mg x (1mmole/16 mg) = 4.25
The empirical formula is therefore,
C(51/4)H34O17/4
C3H8O1
The molar mass of the empirical formula is 60. Therefore, the molecular formula of the compound is,
C9H24O3

5 0

4 years ago

Read 2 more answers

The molar absorptivity of a tyrosine residue at 280 nm is 2000 M-1cm-1, while for tryptophan it is 5500 M-1cm-1. A protein has b

adelina 88 [10]

Answer:

There are 4 tryptophans in the protein.

Explanation:

According to question, protein contains one tyrosine residue and say x number of tryptophans.

Concentration of protein solution = 1.0 micromolar = $1.0\times 10^{-6} Molar$

Molar absorptivity of a protein solution : $\epsilon$

$\epsilon = \epsilon _{tyro}+\epsilon _{tryp}$

$=1\times 2000 M^{-1}cm^{-1}+x\times 5500 M^{-1}cm^{-1}$

Length of the cuvette = l = 1.0 cm

Absorbance of protein solution at 280 nm = A = 0.024

$A=\epsilon \times l\times c$ ( Beer-Lambert's law)

$0.024=(1\times 2000 M^{-1}cm^{-1}+x\times 5500 M^{-1}cm^{-1})\times 1 cm\times 1.0\times 10^{-6} M$

Solving for x :

x = 4

There are 4 tryptophans in the protein.

7 0

3 years ago

. Mitosis allows one cell to grow and split into 2 new cells. Will those 2 new cells split again explain.

Finger [1]

Answer:

Yes

Explanation:

They continue to split and grow and split again until the organism that is carrying them dies.

Sorry I don't really know how to explain:(

4 0

4 years ago

Calculate the energy (in kj/mol) required to remove the electron in the ground state for each of the following one-electron spec

Bess [88]

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

a) Energy change due to transition from n = 1 to n = ∞ ,hydrogen atom .

Z = 1

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0$

Let energy change be E for 1 atom.

$E=E_{\infty}-E_1=0-(-13.6 eV)=13.6 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol$

$E'=1,312.17 kJ/mol$

The energy required to remove the electron in the ground state is 1,312.17 kJ/mol.

b) Energy change due to transition from n = 1 to n = ∞ , $B^{4+}$ atom .

Z = 5

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{5^2}{1^2}eV=-340 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-340eV)=340 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 340eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol$

$E'=32,804.31 kJ/mol$

The energy required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ , $Li^{2+}$ atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-122.4 eV)=122.4 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 122.4 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol$

$E'=11,809.55 kJ/mol$

The energy required to remove the electron in the ground state is 11,809.55 kJ/mol.

d) Energy change due to transition from n = 1 to n = ∞ , $Mn^{24+}$ atom .

Z = 25

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-8,500 eV)=8,500 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol$

$E'=820,107.88 kJ/mol$

The energy required to remove the electron in the ground state is 820,107.88 kJ/mol.

4 0

4 years ago