If the H+ in a solution is 1x10^-1 mol/L what is the OH-

Plants primarily obtain nitrogen from soil in the forms of nitrates (NO₃⁻) and ammonium ions (NH₄), which become available through the processes of nitrogen fixation and nitrification by certain bacteria.

Nitrogen is available to plants mostly in the form of nitrate and ammonium ions in the soil. Atmospheric nitrogen, which composes 78% of the air, is not directly usable by plants due to its strong triple bonds. Through the nitrogen cycle, nitrogen is converted into biologically available forms.

One such process is nitrogen fixation, where free-living or symbiotic bacteria, such as Rhizobium in legume roots, convert gaseous nitrogen (N₂) into ammonia (NH₃). This ammonia can become ammonium (NH₄) and is then available for plant uptake, or it can be further processed by nitrifying bacteria into nitrites (NO₂⁻) and subsequently into nitrates (NO₃⁻), which are absorbed by plant roots.

The availability of ammonia and nitrates in the soil is crucial for plants to synthesize essential compounds like proteins and nucleic acids, making nitrogen a limiting nutrient for plant growth.

You need 1000 milliliters of a 1.0 M AgNO3 solution to provide 169.9 g of pure AgNO₃.

Step 1: Calculate the moles of AgNO₃ needed. To do this, use the given molar mass of AgNO₃, which is 169.87 g/mol.

Moles = mass / molar mass

Moles of AgNO₃ = 169.9 g / 169.87 g/mol = 1.000 mol

Step 2: Determine the volume of the 1.0 M AgNO₃ solution required to obtain 1.000 mol of AgNO₃.

Molarity (M) = moles of solute / liters of solution

Rearranging the formula to solve for volume (liters):

Volume (L) = moles of solute / Molarity

Volume (L) = 1.000 mol / 1.0 M = 1.000 liters

Since the question asks for the volume in milliliters:

Volume (mL) = Volume (L) * 1000

Volume (mL) = 1.000 L * 1000 = 1000 mL

Therefore, you need 1000 milliliters of a 1.0 M AgNO₃ solution to provide 169.9 g of pure AgNO₃.

Let's assume that both He and N₂ have ideal gas behavior.

Then we can use ideal gas law,
     PV = nRT
Where, P is the pressure of gas, V is the volume, n is moles of gas, R is universal gas constant and T is the temperature in Kelvin.

The P and V are same for the both gases.
R is a constant.

The only variables are n and T.

Let's say temperature of He is T₁ and temperature of N₂ is T₂.

n = m/M where n is moles, m is mass and M is molar mass.

Molar mass of He is 4 g/mol and molar mass of N₂ is 28 g/mol

Since mass (m) of both gases are same,
 moles of He = m/4
 moles of N₂ = m/28

Let's apply the ideal gas equation for both gases.
For He gas,
 PV = (m/4)RT₁              (1)

For N₂ gas,
 PV = (m/28)RT₂            (2)

(1) = (2)
(m/4)RT₁ = (m/28)RT₂ 
        T₁/4 = T₂/28
        T₁    = T₂/7
        7T₁  = T₂

Hence, the temperature of N₂ gas is higher by 7 times than the temperature of He gas.

Given the same mass and pressure, the temperature of the helium would be the same as the temperature of nitrogen. This is due to the ideal gas law, which connects the variables of pressure, volume, temperature, and the number of moles of a gas.

The temperature of the helium would be the same as the temperature of nitrogen, given that the quantities (in terms of mass) are the same and they are at the same pressure. This follows from the ideal gas law which states that pressure is directly proportional to temperature and the number of moles of the gas, and inversely proportional to volume. Considering this in context, helium and nitrogen have different molar masses. For a given mass, there will be more moles of helium (which has a lower molar mass) than of nitrogen (which has a higher molar mass). Therefore, if the mass, volume, and pressure are held constant, the temperature for equal masses of the two gases will also be constant. More moles of gas does not mean a higher temperature, contrary to what one might intuitively think.

https://brainly.com/question/30458409

#SPJ3

Types of Bonds can be predicted by calculating the difference in electronegativity.

If, Electronegativity difference is,

                Less than 0.4 then it is Non Polar Covalent

                Between 0.4 and 1.7 then it is Polar Covalent 

                Greater than 1.7 then it is Ionic

For Be and F,

                    E.N of Fluorine          =   3.98

                    E.N of Beryllium        =   1.57

                                                   ________

                    E.N Difference                2.41          (Ionic Bond)

For H and Cl,

                    E.N of Chorine           =   3.16

                    E.N of Hydrogen        =   2.20

                                                   ________

                    E.N Difference                0.96          (Polar Covalent Bond)

For Na and O,

                    E.N of Oxygen          =   3.44

                    E.N of Sodium          =   0.93

                                                       ________

                    E.N Difference                2.51          (Ionic Bond)

For F and F,

                    E.N of Fluorine          =   3.98

                    E.N of Fluorine          =   3.98

                                                        ________

                    E.N Difference                0.00         (Non-Polar Covalent Bond)

Result:

           A polar covalent bond is formed between Hydrogen and Chlorine atoms.

Carbon generally has a higher electronegativity than aluminum, so a carbon atom has a stronger attraction for bonding electrons.

The strength of an atom's attraction for the electrons in a bond is measured by its electronegativity. Among the elements listed, carbon generally has a higher electronegativity than aluminum, which means that an atom of carbon will typically have a stronger attraction for the electrons in a bond. The concept of electronegativity is crucial in understanding the behavior of atoms in different chemical compounds and their ability to attract bonding electrons.

The reaction is
C₂H₄ + O₂ → CO₂ + H₂O

To balance the equation, both side have same number of elements. Here,

In left hand side has                                            in right hand side has

           4 H atoms                                                       2 H atoms 

           2 C atoms                                                       1 C atom 

           2 O atoms                                                       3 O atoms 

First, we have to balance number of C atoms and number of H atoms in both side.  
To balance C atoms, '2' should be added before CO₂ and to balance H atoms, '2' should be added before H₂O. 
Then number of oxygen atoms is 2 x 2 + 2 = 6 in right hand side. So, 3 should be added before O₂ in left hand side.
After balancing the equation should be,

C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

Answer:- P = 43.63% and O = 56.37% and the empirical formula is [tex]P_2O_5[/tex]

Solution:- mass of Phosphorous = 30.98 g

mass of product(phosphorous oxide) = 71.00 g

mass of oxygen = 71.00 - 30.98 = 40.02 g

percentage of phosphorous in the compound = (30.98/71.00)100 = 43.63%

percentage of oxygen in the compound = (40.02/71.00)100 = 56.37%

moles of P = 30.98 g x ( 1mol/30.97 g) = 1.00 mol

moles of O = 40.02 g x (1 mol/ 16.00 g) = 2.50 mol

Ratio of moles of P to O is 1.00:2.50

The whole number ratio is 2:5

So, the empirical formula of the phosphorous oxide formed is [tex]P_2O_5[/tex]

Answer : The percent composition of P = 43.63 % and O = 53.37 % .

Empirical formula of Phosphorous oxide = P₂O₅

Part A : Percent composition :

It is percent of each element present in compound . It is given by formula :

[tex] Percent composition = \frac{mass of element}{ total mass of compound } * 100 [/tex]

Given : Mass of Phosphorous (P ) = 30.98 g

Mass of Compound Phosphorous oxide = 71.00 g

Mass of Oxygen (O) = mass of compound - mass of P

= 71.00 g - 30.98 g = 40.02 g

[tex] Percent composition of P = \frac{mass of P }{ mass of compound} [/tex][tex] = \frac{30.98 g }{71.00 g} * 100 [/tex]

Percent composition of P = 43.63 %

[tex] Percent composition of O = \frac{mass of Al}{Mass of compound } * 100
= \frac{40.02 g }{71.00 g} * 100 [/tex]

Percent composition of O = 53.37 %

Part B : Empirical formula of Phosphorous Oxide.

Empirical formula is formula which shows the proportion of element present in a compound . Following are the steps to calculate empirical formula of an compound :

Step 1 : Find masses of each element .

Mass of P = 30.98 g

Mass of O = 40.02 g

Step 2 : Conversion of masses of element to its mole .

[tex] Mole = \frac{mass }{molar mass } [/tex]

Given : Mass of P = 30.98 g [tex] Molar mass of P = 30.97 \frac{g}{ mol} [/tex]

[tex] Mole of P = \frac{30.98 g }{30.97\frac{g}{mol} } = 1 mol [/tex]

Given: Mass of O = 40.02 g [tex] Molar mass of O = 15.99\frac{g}{mol} [/tex]

[tex] Mole of O = \frac{40.02 g}{15.99\frac{g}{mol}} = 2.5 mol [/tex]

Step 3 : Finding Ratio of mole .

In this step ratio is found by dividing each mole by smallest mole .Since mole of P is smaller , so this will be used for division.

[tex] Ratio of mole of P= \frac{mole of P }{mole of P } = \frac{1 mol}{1 mol} = 1 [/tex]

[tex] Ratio of mole of O = \frac{ mole of O }{mole of P } = \frac{2.5 }{1 } = 2.5 [/tex]

Hence , Ratio of P : O = 1 : 2.5

Since the ratio is in fraction , it need to be converted to whole number . So we multiply the ratio by such a minimum number which gives us a whole number ratio. This step is skipped if the ratio already comes in whole number.

On multiplication the ratio by 2 :

Ratio of P : O = ( 1 : 2.5 ) * 2 = 1 * 2 : 2.5 * 2

Ratio of P : O = 2 : 5

Step 4 : Writing the empirical formula

The ratio of P and O is 2: 5 , which gives the empirical formula of compound as P₂O₅ .

Final answer:

J. J. Thomson discovered electrons, Ernest Rutherford found protons, and James Chadwick discovered neutrons, all of which conflicted with Dalton's theory of indivisible atoms.

Explanation:

Dalton's Atomic Theory stated that matter is made up of indivisible particles called atoms. However, subsequent scientists' discoveries supported or differed from Dalton's theory in the following ways:

Answer:

opposite ions attract

Saturated hydrocarbons, such as alkanes, are utilized in combustion processes for producing heat due to their single carbon-to-carbon bonds.

Saturated hydrocarbons, also known as alkanes, involve the direct use of hydrocarbons with only single carbon-to-carbon bonds. An example of an activity that most likely involves the direct use of saturated hydrocarbons is combustion processes for producing heat. These hydrocarbons are composed entirely of single bonds and are saturated with hydrogen.

Answer:

1.67m

Explanation:

Factors affecting the rate of a chemical reaction in the context of a specific reaction equation.

The factors that affect the rate of the reaction Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g) include:

Catalyst: Adding a catalyst can increase the rate of the reaction.

Surface area: Grinding the zinc solid into a fine powder increases its surface area, enhancing the reaction rate.

Concentration and Temperature: Increasing the concentration of HCl or raising the temperature can also speed up the reaction.

The molecule containing a special group attached to its side chain is a fundamental group. The molecules are acid, ether, ester, aldehyde, ketone, amine, and alcohol.

Functional groups are the specialized molecules that are attached to the main chain and vary the chemical and physical properties of the compound. The molecule containing hydroxide is acid, and the single oxygen atom is ether.

Compound with double-bonded oxygen to alkyl or aromatic group is ester, carbonyl group on carbon with hydrogen and R group is an aldehyde, carbonyl group attached to carbon-containing two alkyl or aromatic groups is a ketone.

The Amine group includes an aromatic or an alkyl group attached to two hydrogens of the nitrogen atom. The alcohol group is the attachment of the -OH group to a carbon-containing hydrogen and R group.

Learn more about the functional groups here:

https://brainly.com/question/14618322

#SPJ2

Answer:

[tex]SO_4^-[/tex]

Explanation:

The sodium sulfate compound corresponds to a water soluble salt.

When a salt dissolves it will dissociate into the ions that compose it.

It dissociates into Cations (positively charged ions) and anions (negatively charged ions)

Sodium sulfate formula

[tex]Na_2SO_4[/tex]

Sodium sulfate when dissolved in water

[tex]Na_2SO_4\stackrel{H_2O}{\longrightarrow} 2Na^+SO_4^-[/tex]

Ions present in the solution:

[tex]Na^+[/tex] Cation

[tex]SO_4^-[/tex] Anion

Answer:It might have 10 neutrons

Reason: This is because an isotope of the same element must have the same number of electrons and protons so it can be assumed that the amount of neutrons can be changed.