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2 years ago

A human hair is 75 um across. How many inches is this?

Chemistry

2 answers:

jonny [76]2 years ago

7 0

Answer:

0.003in

Explanation:

75 what exactly

Nikitich [7]2 years ago

7 0

Answer:

2.9528e-6

Explanation:

or 0.0029

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From the following list of elements, those that will always form ionic compounds in a 1:2 ratio with zinc.

harina [27]

Answer:

A. iodine

C. fluorine

F. bromine

Explanation:

Ionic bonds occur mostly between metals and non-metals. Usually, a wide electronegativity difference is preferred between the two atoms. This makes one atom more desirous to gain electron and other more willing to donate electrons.

To have Zn forming a compound in the ratio of 1 to 2, the combining power must be similar to this.

The dominant oxidation state of Zn is the is the +2 state.

The other combining atoms must have the ability to recieve the two electrons.

The halogens fit perfectly into this picture. They need just an electron to attain nobility. They are also highly electronegative. If two halogens combines with the Zn, then the ionic bond will result.

The halogens are fluorine,chlorine, bromine, iodine and astatine.

They will form these compounds:

ZnF₂, ZnBr₂ and ZnI₂

3 0

2 years ago

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute th

MrRissso [65]

The question is incomplete, complete question is :

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.

$\frac{K_1}{K_2}=?$

Activation energy of the reaction 1 , $Ea_1$ = 14.0 kJ/mol

Activation energy of the reaction 2, $Ea_1$ = 11.9 kJ/mol

Answer:

0.4284 is the ratio of the rate constants.

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

The expression used with catalyst and without catalyst is,

$\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}$

$\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}$

where,

$K_2$ = rate constant reaction -1

$K_1$ = rate constant reaction -2

Activation energy of the reaction 1 , $Ea_1$ = 14.0 kJ/mol = 14,000 J

Activation energy of the reaction 2, $Ea_1$ = 11.9 kJ/mol = 11,900 J

R = gas constant = 8.314 J/ mol K

T = temperature = $25^oC=273+25=298 K$

Now put all the given values in this formula, we get

$\frac{K_1}{K_2}=e^{\frac{11,900- 14,000Jl}{8.314 J/mol K\times 298 K}}=2.3340$

0.4284 is the ratio of the rate constants.

7 0

2 years ago

Glade air freshener gel “disappearing” is an example of

netineya [11]

Answer:

Vaporization

Explanation:

Vaporization is the change of a specie to the gaseous state. To 'disappear' in this case simply means to change to the gaseous state.

Substances with high vapour pressure tend to be easily converted to vapour phase. Hence if Glade air freshener gel 'disappears' easily, then it has a high vapour pressure and is easily converted to vapour (gas).

8 0

1 year ago

A solution has a concentration of 0.6mol/dm³. If a container of this solution holds 3 moles of solute, what volume of solution i

FinnZ [79.3K]

Answer: 3 moles solute x 1 dm^3/0.60 moles solute = 5 dm^3

Explanation:

6 0

1 year ago

Sodium reacts with chlorine gas according to the following reaction: 2Na(s)+Cl2(g)→2NaCl(s) What volume of Cl2 gas, measured at

asambeis [7]

Answer:6.719Litres of Cl2 gas.

Explanation:According to eqn of rxn

2Na +Cl2=2NaCl

P=689torr=689/760=0.91atm

T=39°C+273=312K

according to stoichiometry of the reaction,1Moles of Cl2 gives 2moles of NaCl

But 28g of NaCl was given,we have to convert this to moles by using the relation, n=mass/MW

MW of NaCl=23+35.5=58.5g/mol

n=28g(mass given of NaCl)/58.5

n=0.479moles of NaCl

Going back to the reaction,

if 1moles of Cl2 produces 2moles of NaCl

x moles of Cl2 will give 0.479moles of NaCl.

x=0.479*1/2

x=0.239moles of Cl2.

To find the volume, we use ideal ggas eqn,PV=nRT

V=nRT/P

V=0.239*0.082*312/0.91

V=6.719Litres

6 0

1 year ago