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djverab [1.8K]
3 years ago
7

Why is it necessary to obtain 25 to 50 hairs from this woman?

Chemistry
1 answer:
Vedmedyk [2.9K]3 years ago
5 0

To get the correct data

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What is the mass in grams of 1.50 X 1023 molecules of NaOH?
avanturin [10]

Answer:

a. 10.0 g

Explanation:

The molar mass of NaOH is 39.997 grams which was calculated by adding the molar mass of Sodium, Oxygen, and Hydrogen 22.990 + 15.999 + 1.008 = 39.997. But before we do anything with this number we must first convert 1.50 * 10^{23} molecules to moles. We can do this by dividing by Avogadro's number or 6.022*10^{23}  ( \frac{1.50*10^{23}}{6.022*10^{23}} = 0.24909 moles). We then convert moles to grams by multiplying by the molar mass of NaOH, (0.24909 * 39.997 = 9.96 grams). This is almost exactly 10 and the margin of error is likely due to the original numbers not being precise enough.

5 0
4 years ago
Compare to the sample of helium at STP, the same sample of helium at a higher temperature and a lower pressure.
Molodets [167]

Here we have to compare the state of helium gas at STP and high temperature and low pressure.

At STP (standard condition of temperature and pressure) i.e. 273K temperature and 1 bar pressure. At STP helium gas will behave as a real gas.

At higher temperature and low pressure Helium will behave as an ideal gas.

The ideal gas conditions are developed on taking into account two factors: (i) the gas molecules are point of mass and having no volume. (ii) there is no existence of force of attraction between the molecules.

The deviation from ideal gas to the real gas depends upon the van der waals' interaction between the gas molecules. Now in low pressure and high temperature, we can ignore the volume and also the inter-molecular force of attraction. Thus the gas sample can behaved as ideal gas.

But at elevated pressure and low temperature i.e. STP the assumptions are not valid and it will behave as real gas.    

4 0
3 years ago
A solution of HCOOH has 0.16M HCOOH at equilibrium. The Ka for HCOOH is 1.8×10−4. What is the pH of this solution at equilibrium
kotykmax [81]

Answer:

pH=2.28

Explanation:

Hello,

In this case, for the acid dissociation of formic acid (HCOOH) we have:

HCOOH(aq)\rightarrow H^+(aq)+HCOO^-(aq)

Whose equilibrium expression is:

Ka=\frac{[H^+][HCOO^-]}{[HCOOH]}

That in terms of the reaction extent is:

1.8x10^{-4}=\frac{x*x}{0.16-x}

Thus, solving for x which is also equal to the concentration of hydrogen ions we obtain:

x=0.00528M

[H^+]=0.00528M

Then, as the pH is computed as:

pH=-log([H^+])

The pH turns out:

pH=-log(0.00528M)\\\\pH=2.28

Regards.

7 0
4 years ago
I neeeeeeed helpppppppp
tresset_1 [31]

Answer:

I cAnT sEe It

Explanation:

like fr i cant see anything on that picture

5 0
3 years ago
What's autonization of water​
GarryVolchara [31]

Explanation:

The self-ionization of water (also autoionization of water, and autodissociation of water) is an ionization reaction in pure water or in an aqueous solution, in which a water molecule, H2O, deprotonates (loses the nucleus of one of its hydrogen atoms) to become a hydroxide ion, OH−. The hydrogen nucleus, H+, immediately protonates another water molecule to form hydronium, H3O+. It is an example of autoprotolysis, and exemplifies the amphoteric nature of water

Animation of the self-ionization of water

Chemically pure water has an electrical conductivity of 0.055 μS/cm. According to the theories of Svante Arrhenius, this must be due to the presence of ions. The ions are produced by the water self-ionization reaction, which applies to pure water and any aqueous solution:

H2O + H2O ⇌ H3O+ + OH−

Expressed with chemical activities a, instead of concentrations, the thermodynamic equilibrium constant for the water ionization reaction is:

{\displaystyle K_{\rm {eq}}={\frac {a_{\rm {H_{3}O^{+}}}\cdot a_{\rm {OH^{-}}}}{a_{\rm {H_{2}O}}^{2}}}}

which is numerically equal to the more traditional thermodynamic equilibrium constant written as:

{\displaystyle K_{\rm {eq}}={\frac {a_{\rm {H^{+}}}\cdot a_{\rm {OH^{-}}}}{a_{\rm {H_{2}O}}}}}

under the assumption that the sum of the chemical potentials of H+ and H3O+ is formally equal to twice the chemical potential of H2O at the same temperature and pressure.[1]

Because most acid–base solutions are typically very dilute, the activity of water is generally approximated as being equal to unity, which allows the ionic product of water to be expressed as:[2]

{\displaystyle K_{\rm {eq}}\approx a_{\rm {H_{3}O^{+}}}\cdot a_{\rm {OH^{-}}}}

In dilute aqueous solutions, the activities of solutes (dissolved species such as ions) are approximately equal to their concentrations. Thus, the ionization constant, dissociation constant, self-ionization constant, water ion-product constant or ionic product of water, symbolized by Kw, may be given by:

{\displaystyle K_{\rm {w}}=[{\rm {H_{3}O^{+}}}][{\rm {OH^{-}}}]}

where [H3O+] is the molarity (≈ molar concentration)[3] of hydrogen or hydronium ion, and [OH−] is the concentration of hydroxide ion. When the equilibrium constant is written as a product of concentrations (as opposed to activities) it is necessary to make corrections to the value of {\displaystyle K_{\rm {w}}} depending on ionic strength and other factors (see below).[4]

At 25 °C and zero ionic strength, Kw is equal to 1.0×10−14. Note that as with all equilibrium constants, the result is dimensionless because the concentration is in fact a concentration relative to the standard state, which for H+ and OH− are both defined to be 1 molal (or nearly 1 molar). For many practical purposes, the molal (mol solute/kg water) and molar (mol solute/L solution) concentrations can be considered as nearly equal at ambient temperature and pressure if the solution density remains close to one (i.e., sufficiently diluted solutions and negligible effect of temperature changes). The main advantage of the molal concentration unit (mol/kg water) is to result in stable and robust concentration values which are independent of the solution density and volume changes (density depending on the water salinity (ionic strength), temperature and pressure); therefore, molality is the preferred unit used in thermodynamic calculations or in precise or less-usual conditions, e.g., for seawater with a density significantly different from that of pure water,[3] or at elevated temperatures, like those prevailing in thermal power plants.

We can also define pKw {\displaystyle \equiv } −log10 Kw (which is approximately 14 at 25 °C). This is analogous to the notations pH and pKa for an acid dissociation constant, where the symbol p denotes a cologarithm. The logarithmic form of the equilibrium constant equation is pKw = pH + pOH.

7 0
3 years ago
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