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3 years ago

3.75 g of an unknown gas at 59 °C and 1.00 atm is stored in a 1.35-L flask. What is the molar mass of the gas?

1 answer:

4 0

You need to find moles of the gas, so you would use the ideal gas law:
PV=nRT
Pressure
Volume
n=moles
R= gas constant
Tenperature in Kelvin
n= PV/RT
(1.00atm)(1.35L)/(.08206)(332K) = 0.050mol
Molar mass is grams per mole, so
(3.75g/.050mol) = 75g/mol

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Can someone please help I need help?!?!

anzhelika [568]

Answer:

a: 6 moles

b: 6.75 moles

c: 5 grams

Explanation:

a: mole ratio 2:3

b: mole ratio 2:3

c: mole ratio 2:2

7 0

3 years ago

What is the linkage that binds one monosaccharide to another?

ra1l [238]

Answer:

hope this helps

Explanation:

glycosidic bond

A covalent bond formed between a carbohydrate molecule and another molecule (in this case, between two monosaccharides) is known as a glycosidic bond (Figure 4). Glycosidic bonds (also called glycosidic linkages) can be of the alpha or the beta type.

4 0

2 years ago

Read 2 more answers

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

Using the periodic tables, identify the heaviest member of each of the following groups:

Lady_Fox [76]

Answer:

(a) Alkali metals: Francium (Fr)

(b) Chalcogens: Polonium (Po)

(c) Noble gases: Radon (Rn)

(d) Alkaline earth metals: Radium (Ra)

Explanation:

In the periodic table, the atomic mass increases down the group. Therefore, the last element of a group is the heaviest element of the group.

(a) alkali metals: The chemical elements that are present in group 1 of the periodic table, except hydrogen.

The heaviest member of this group is francium (Fr)

(b) chalcogens: The chemical elements that are present in group 16 of the periodic table

The heaviest member of this group is polonium (Po)

(c) noble gases: The chemical elements that are present in group 18 of the periodic table

The heaviest member of this group is radon (Rn)

(d) alkaline earth metals: The chemical elements that are present in group 2 of the periodic table.

The heaviest member of this group is radium (Ra)

7 0

3 years ago

For an aqueous solution of sodium chloride (NaCl) .Determine the molarity of 3.45L of a solution that contains 145g of sodium

worty [1.4K]

Molar mass NaCl = 58.44 g/mol

number of moles:

mass NaCl / molar mass

145 / 58.44 => 2.4811 moles of NaCl

Volume = 3.45 L

Therefore :

M = moles / volume in liters:

M = 2.4811 / 3.45

M = 0.719 mol/L⁻¹

hope this helps!

6 0

3 years ago