What are complementary bases?

Dilution of a buffer solution with a small amount of water will ________ the ph of the solution.

Daniel [21]

The Dilution of a buffer solution with a small amount of water will raise the PH of the solution. Dilution is the process of "reducing the concentration of a solute in a solution by simply adding more solvent to the solution, such as water." Dilution involves increasing the amount of solvent in a solution without increasing the amount of solute.

When a solvent is added, the concentration of the solute in the solution is diluted. By removing the solvent, concentration raises the solute concentration in the solution. Dilution is the process of lowering a given solute's concentration in its solution.

To learn more about Dilution, click here.

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4 0

2 years ago

Calculate the mass of H2OH2O produced by metabolism of 1.6 kgkg of fat, assuming the fat consists entirely of tristearin (C57H11

kakasveta [241]

Answer: 1779.8g of H2O

Explanation:

C57H110O6 + O2 —> CO2 + H20

To calculate the amount of water(H2O) produced by the reaction of 1.6Kg tristearin(C57H110O6) and O2, we must first balance the equation.

To do so, add 2 in front of C57H110O6, 163 in front of O2, 114 in front of CO2 and 110 in front of H2O i.e

2C57H110O6 + 163O2 —> 114CO2 + 110H20

Next, we have to calculate the mass of tristearin(C57H110O6) that reacted and the mass of water produced from the balanced equation. This can be done by:

MM of C57H110O6 = (57x12) + (110x1) + (6x16) = 684 + 110 + 96 = 890g/mol

Mass conc of C57H110O6 = 890 x 2 = 1780g

MM of H2O = (2x1) + 16 = 18g/mol

Mass conc of H2O = 110 x 18 = 1980g

Next, we will find the amount of water produced by 1.6Kg(i.e 1600g) of tristearin(C57H110O6). This is done by:

From the equation,

1780g of C57H110O6 produced 1980g of H2O.

Therefore 1600g of C57H110O6 will produce = (1600x1980)/1780 = 1779.8g of H2O

5 0

3 years ago

Solid to Gas

miskamm [114]

Answer:

c. c. c c c c c cçcc. ccccccccc

Explanation:

ccc cc

7 0

3 years ago

Ammonia (NH3) is one of the most common chemicals produced in the united states. it is used to make fertilizer and other product

Greeley [361]

Answer:

a) N2 is the limiting reactant

b) 1215.9 grams NH3

c) 2283.6 grams H2

Explanation:

Step 1: Data given

Mass of N2 = 1000 grams

Molar mass N2 = 28 g/mol

Mass of H2 = 2500 grams

Molar mass H2 = 2.02 g/mol

Step 2: The balanced equation

N2(g) +3H2(g) → 2NH3(g)

Step 3: Calculate moles N2

Moles N2 = mass N2 / molar mass N2

Moles N2 = 1000 grams / 28.0 g/mol

Moles N2 = 35.7 moles

Step 4: Calculate moles H2

Moles H2 = 2500 grams / 2.02 g/mol

Moles H2 = 1237.6 moles

Step 5: Calculate limiting reactant

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

N2 is the limiting reactant. There will be consumed 35.7 moles.

H2 is in excess. There will react 3*35.7 = 107.1 moles

There will remain 1237.6 - 107.1= 1130.5 moles H2

This is 1130.5 * 2.2 = 2283.6 grams H2

Step 6: Calculate moles NH3

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

For 35.7 moles N2 we'll have 2*35.7 = 71.4 moles NH3

Step 7: Calculate mass NH3

Mass NH3 = moles NH3 * molar mass NH3

Mass NH3 = 71.4 moles * 17.03 g/mol

Mass NH3 = 1215.9 grams NH3

7 0

3 years ago

Give the formulas of the compounds in each set (a) lead(l|) oxide and lead(lV) oxide; (b) lithium nitride, lithium nitrite, and

Ivanshal [37]

Answer:

For a: The formula of lead (II) oxide and lead (IV) oxide is $PbO\text{ and }PbO_2$ respectively.

For b: The formula of lithium nitride, lithium nitrite and lithium nitrate is $Li_3N,LiNO_2\text{ and }LiNO_3$ respectively.

For c: The formula of Strontium hydride and strontium hydroxide is $SrH_2\text{ and }Sr(OH)_2$ respectively.

For d: The formula of magnesium oxide and manganese (II) oxide is $MgO\text{ and }MnO$ respectively.

Explanation:

All the given compounds are ionic compounds. This means that between the atoms complete sharing of electrons takes place.

For a:

Lead is the 82th element of periodic table having electronic configuration of $[Xe]4f^{14}5d^{10}6s^26p^2$ .

To form $Pb^{2+}$ ion, this element will loose 2 electrons and to form $Pb^{4+}$ ion, this element will loose 4 electrons.

Oxygen is the 8th element of periodic table having electronic configuration of $[He]2s^22p^4$ .

To form $O^{2-}$ ion, this element will gain 2 electrons.

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for lead (II) oxide is PbO and for lead (IV) oxide is $PbO_2$

Thus, the formula of lead (II) oxide and lead (IV) oxide is $PbO\text{ and }PbO_2$ respectively.

For b:

Lithium is the 3rd element of periodic table having electronic configuration of $[He]2s^1$ .

To form $Li^{+}$ ion, this element will loose 1 electron.

Nitrogen is the 7th element of periodic table having electronic configuration of $[He]2s^22p^3$ .

To form $N^{3-}$ ion, this element will gain 3 electrons.

Nitrite ion is a polyatomic ion having chemical formula of $NO_2^{-}$

Nitrate ion is a polyatomic ion having chemical formula of $NO_3^{-}$

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for lithium nitride is $Li_3N$ , for lithium nitrite is $LiNO_2$ and for lithium nitrate is $LiNO_3$

Thus, the formula of lithium nitride, lithium nitrite and lithium nitrate is $Li_3N,LiNO_2\text{ and }LiNO_3$ respectively.

For c:

Strontium is the 38th element of periodic table having electronic configuration of $[Kr]5s^2$ .

To form $Sr^{2+}$ ion, this element will loose 2 electrons.

Hydrogen is the 1st element of periodic table having electronic configuration of $1s^1$ .

To form $H^{-}$ ion, this element will gain 1 electron and is named as hydride ion.

Hydroxide ion is a polyatomic ion having chemical formula of $OH^{-}$

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for strontium hydride is $SrH_2$ and for strontium hydroxide is $Sr(OH)_2$

Thus, the formula of Strontium hydride and strontium hydroxide is $SrH_2\text{ and }Sr(OH)_2$ respectively.

For d:

Magnesium is the 12th element of periodic table having electronic configuration of $[Ne]3s^2$ .

To form $Mg^{2+}$ ion, this element will loose 2 electrons.

Manganese is the 25th element of the periodic table having electronic configuration of $[Ar]3d^54s^2$

To form $Mn^{2+}$ ion, this element will loose 2 electrons.

Oxygen is the 8th element of periodic table having electronic configuration of $[He]2s^22p^4$ .

To form $O^{2-}$ ion, this element will gain 2 electrons.

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for magnesium oxide is MgO and for manganese (II) oxide is MnO.

Thus, the formula of magnesium oxide and manganese (II) oxide is $MgO\text{ and }MnO$ respectively.

7 0

3 years ago