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3 years ago

Consider the reaction. mc015-1.jpg How many grams of N2 are required to produce 100.0 L of NH3 at STP?

1 answer:

Nadusha1986 [10]3 years ago

3 0

The grams of N2 that are required to produce 100.0 l of NH3 at STP

At stp 1moles = 22.4 l. what about 100.0 L of NH3

= 100 / 22.4 lx1 moles = 4.46 moles of NH3

write the reacting equation

N2+3H2 =2NH3
by use of mole ratio between N2 to NH3 which is 1:2 the moles of N2 =4.46/2 =2.23 moles of N2

mass = moles x molar mass

= 2.23moles x 28 g/mol = 62.4 grams

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35. In the collision theory, a collision that leads to the formation of products is called an

FinnZ [79.3K]

Answer:

It's Effective Collision.

Explanation:

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7 0

3 years ago

A compound consisting of B, N, and H undergoes elemental analysis. The % composition by mass is found to be 40.28% B, 52.20% N,

swat32

Empirical formula is the simplest ratio of components making up a compound.

The percentage composition of each element has been given

therefore the mass present of each element in 100 g of compound is

B N H

mass 40.28 g 52.20 g 7.53 g

number of moles

40.28 g / 11 g/mol 52.20 g / 14 g/mol 7.53 g / 1 g/mol

= 3.662 mol = 3.729 mol = 7.53 mol

divide the number of moles by the least number of moles, that is 3.662

3.662 / 3.662 3.729 / 3.662 7.53 / 3.662

= 1.000 = 1.018 = 2.056

the ratio of the elements after rounding off to the nearest whole number is

B : N : H = 1 : 1 : 2

therefore empirical formula for the compound is B₁N₁H₂

that can be written as BNH₂

3 0

3 years ago

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

3 years ago

Find How many joules of heat energy are needed to raise the temperature of 10grams of Aluminum from 20*C to 40*C. If the specifi

dmitriy555 [2]

Explanation:

the answer and the working out is shown above, hope it helps.

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2 years ago

How is an isotope different from the standard form of a chemical element?

velikii [3]

Answer:

The standard form of a chemical element is the natural mixture of several isotopes of the same element, which is atoms with the same number of protons but different number of neutrons, while an isotope is a particular kind of atom with a definite number of neutrons.

Explanation:

A <em>chemical element</em> is a pure substance formed by atoms with the same atomic number (number of protons). This is because it is the number of protons what identifies an element.

For example: oxygen is a chemical element, so oxygen is formed by only atoms of oxygen, and the atomic number of those atoms is 8, because every oxygen atom has 8 protons.

Nevertheless, some atoms of oxygen, may have different number of neutrons. Isotopes are different kind of atoms of the same element, which only differ in the number of neutrons. So, some atoms of oxygen will have 8 neutrons, other 9 neutrons, and other 10 neutrons (those are the stable isotopes of oxygen).

That difference in neutrons, is generally accepted that, does not modifiy substantially the chemical properties of the element, but the mass number. So, the isotopes with more neutrons wil be heavier, and the isotopes with less neutrons will be lighter.

Mass number = number of protons + number of neutrons.

In general a chemical element is formed by a mixutre of isotopes of the same element.

4 0

3 years ago

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