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algol13
3 years ago
12

What happens to iron oxide during decomposition

Chemistry
1 answer:
elena55 [62]3 years ago
4 0

Answer:

it gets reduced from a +3 oxidation to a 0.

Explanation:

the decomposition of iron oxide to elemenTal can be represented by the following equation: iron oxide (Fe203), the oxidation state of iron is +3 while that of oxygen is -2. therefore, the above reaction is a redox (reduction oxidation reaction)

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Caculate the mass of oxygen needed to burn 175.0grams of octane
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A water wave traveling in a straight line on a lake is described by the equation
y(x,t)=(3.75cm)cos(0.450cm?1x+5.40s?1t)
where y is the displacement perpendicular to the undisturbed surface of the lake.
8 0
4 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
Regla de kernel en hg<br> Quimica
malfutka [58]
Uhm Did u have Questions or pic. So I can Answer your Question
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6 0
3 years ago
A student reacts 20.0 grams of sodium hydroxide with excess hydrochloric acid, and obtains a
Tcecarenko [31]

Answer:

Percent yield of NaCl = 87.75 %

Explanation:

Data given

mass of sodium = 20.0 g

actual yield of Sodium chloride = 25 g

percent yield of sodium chloride = ?

Reaction Given:

              NaOH + HCI --> NaCl + H₂O

Solution:

First we have to find theoretical yield.

So, Look at the reaction

                   NaOH + HCI ---------> NaCl + H₂O

                    1 mol                           1 mol

As 1 mole of NaOH give 1 mole of NaCl

Convert moles to mass

molar mass of NaOH = 23 + 16 + 1 = 40 g/mol

molar mass of NaCl = 23 +35.5 = 58.5 g/mol

Now

             NaOH     +    HCI    --------->   NaCl      +    H₂O

           1 mol (40 g/mol)               1 mol (58.5 g/mol)

                40 g                                     58.5 g

So 40 g of NaOH gives 58.5 g of NaCl so how many grams of Sodium chloride will be produced by 20 g of NaOH.

Apply Unity Formula

                  40 g of NaOH ≅ 58.5 g of NaCl

                  20 g of NaOH ≅ X g of NaCl

Do cross multiply

               g of NaCl =  58.5 g x 20 g / 40 g

               g of NaCl = 29.25 g

So the Theoretical yield of NaCl = 29.25 g

Now Find the percent yield of NaCl

Formula Used

          percent yield = actual yield /theoretical yield x 100 %

Put value in the above formula

           percent yield = 25 / 29.2 x 100 %

           percent yield = 87.75 %

percent yield of NaCl = 87.75 %

7 0
4 years ago
The mass of gemstones and pearls is usually expressed in units called carats. One carat is exactly equal to 200 mg. What is the
lys-0071 [83]

Answer: The mass in grams for a 9.50 carat diamond is 1.9.

Explanation:

Given :

The mass of gemstones and pearls is usually expressed in units called carats.

we have to find mass in grams for 9.50 carat diamond.

1 carat = 200 mg

Thus 9.50 carat=\frac{200}{1}\times 9.50=1900mg

Also 1 mg =10^{-3}g

Thus  1900mg =\frac{10^{-3}}{1}\times 1900=1.9g

Thus the mass in grams for a 9.50 carat diamond is 1.9

8 0
4 years ago
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