Which of these is a base?

Organize the following list of atoms from lowest to highest Electronegativity:

Anon25 [30]

Answer:

Barium<Strontium<Calcium <Magnesium< Beryllium

Explanation:

Electronegativity is defined as the ability of an atom in a bond to attract the shared electrons of the bond towards itself.

Electronegativity is a periodic trend that decreases down the group and increases across the period.

Hence, if i want to arrange Beryllium, Barium, Strontium, Magnesium, Calcium in order of increasing electronegativity, i will have;

Barium<Strontium<Calcium <Magnesium< Beryllium

5 0

3 years ago

Jai besoi daide

JulijaS [17]

Answer:

i dont know what you are saying

Explanation:

????????????

4 0

2 years ago

Please help!!!!!!!!!!!!!!!!!!!

zimovet [89]

Answer:

Your body cells use the oxygen you breathe to get energy from the food you eat. This process is called cellular respiration. During cellular respiration the cell uses oxygen to break down sugar. Breaking down sugar produces the energy your body needs.

Explanation:

Brainly pls

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:

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

3 years ago

a book with a mass of 1 kg is dropped from a height of 3 m. what is the potential energy of the book when it reaches the floor?

mafiozo [28]

Well, first of all, the formula for finding potential energy is;
PE=mgh
Where; m is the mass
g is the gravitational force or acceleration due to gravity
h is the height.
Anyway, according to the question, the mass is 1kg, the acceleration due to gravity has a constant value of 10ms² . And the height is 3m. Now you just have to use all these in the formula. So;
mgh= 1 x 10 x 3. That will be 30. And the unit of potential energy is Joule. So the answer is 30 joules. Hope i helped. Have a nice day.

5 0

3 years ago