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GrogVix [38]
3 years ago
9

The gases listed are at standard temperature and pressure: · 1 mol of CO2 gas · 1 mol of N2 gas · 2 mol of O2 gas Which statemen

t is correct? The volume of 1 mol of CO2 is greater than that of 2 mol of O2 . The volume of 1 mol of CO2 is greater than that of 1 mol of N2 . The volumes of 2 mol of O2 and 1 mol of N2 gases are the same. **The volumes of 1 mol of CO2 and 1 mol of N2 gases are the same.**
Chemistry
2 answers:
andrezito [222]3 years ago
8 0
The right answer for the question that is being asked and shown above is that: "The volumes of 1 mol of CO2 and 1 mol of N2 gases are the same." The <span>statement that is correct as far as the </span>gases listed are at standard temperature and pressure is concerned is that t<span>he volumes of 1 mol of CO2 and 1 mol of N2 gases are the same</span>
Elena L [17]3 years ago
6 0

Answer : The volume of 1 mole of CO_{2} and N_{2} gases are the same.

Solution: Given,

Moles of CO_{2} = 1 mole

Moles of N_{2} = 1 mole

Moles of O_{2} = 2 mole

Formula used for ideal gas is :

P V = n R T

According to the question, the gases are at standard temperature and pressure. So, the volume of gases only depends on the number of moles. This means that the higher the number of moles, higher will be the volume of gas.

The moles of O_{2} are more than the moles of CO_{2} and N_{2}. So, the volume of O_{2} will be more.

And the moles of CO_{2} and N_{2} are equal. Therefore, their volumes are also equal.

Therefore, the best option is the volume of 1 mole of CO_{2} and N_{2} gases are the same.

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Ionic bonds are formed when one of the two atoms that are reacting has excess electrons and transfer the electrons to the atom that is deficient in electrons. During the formation of the ionic bond, one of the reacting atoms will donate electrons and form positive ion.
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Calculate the mass of bromine in 50.0 g of potassium bromide
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Answer:

6 Percent Composition. 1. Molar Mass of KBr K = 1(39.10) = 39.10 Br =1(79.90) =79.90 MM = 119.0 79.90 g        119.0 g = 0.6714 3. 0.6714 x 50.0g = 33.6 g Br 2.

4 0
3 years ago
Consider the chemical equation below. Upper M g (upper O upper H) subscript 2 plus 2 upper H upper N upper O subscript 3 right a
masha68 [24]

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Upper H superscript plus, plus upper O upper H superscript minus right arrow upper H subscript 2 upper O.

Explanation:

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An alloy of bronze is manufactured by melting 51.2 g of copper with 6.84 g of tin. What is the percent copper in the bronze?
Harlamova29_29 [7]

Answer:

Percentage of copper = 88%

Explanation:

Given data:

Mass of copper = 51.2 g

Mass of tin = 6.84 g

Percentage of copper = ?

Solution:

Formula:

Percentage of copper = mass of copper / total mass × 100

Now we will determine the total mass:

Total mass = mass of copper + mass of  tin

Total mass = 51.2 g + 6.84 g

Total mass = 58.04 g

Now we will calculate the percentage of copper.

Percentage of copper = 51.2 g / 58.04 g × 100

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5 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:

1)

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}.

2)

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.

3.

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
4 years ago
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