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3 years ago

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air hockey is a game that you play on the specially-designed table each player tries to slide a pug cross the table into a go us

ing mileage the table has tiny holes that create cushion of air that lives the focus more distance about the playing surface the cushion of air increases the speed of the puck because it's......mm

1 answer:

AURORKA [14]3 years ago

5 0

Answer:

i don't know man really my bad

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Butane (c4h10) + oxygen -> carbon dioxide + water

ra1l [238]

Please include more description <span />

4 0

3 years ago

The half-life of phosphorus-32 is 14.30 days. how many milligrams of a 20.00 mg sample of phosphorus-32 will remain after 85.80

ZanzabumX [31]

The amount of sample that is left after a certain period of time, given the half-life, h, can be calculated through the equation.

A(t) = A(o) (1/2)^(t/d)

where t is the certain period of time. Substituting the known values,

A(t) = (20 mg)(1/2)^(85.80/14.30)

Solving,

A(t) = 0.3125 mg

Hence, the answer is 0.3125 mg.

7 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 balanced chemical equation is shown below

IrinaVladis [17]

Answer:

The answer is A.

Explanation:

C6H12O6 + 6O2 -> 6CO2 + 6H20

C6H12O6 + 6O2 -> 6CO2 + 6H20At the reactant side, there are 6 carbon atoms, 12 hydrogen atoms and 18 oxygen atoms.

atoms, 12 hydrogen atoms and 18 oxygen atoms.At product side, there are 6 carbon atoms, 12 hydrogen atoms and 18 oxygen atoms.

(In a balanced equation, the no. of atoms on the reactant side must be <em><u>e</u></em><em><u>q</u></em><em><u>u</u></em><em><u>a</u></em><em><u>l</u></em><em> </em>to the no. of atoms on the product side)

4 0

3 years ago

A student dissolved 5.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution

Marianna [84]

Answer:

0.136g

Explanation:

A student dissolved 5.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution and then diluted it with water to give 275. mL of a final solution. How many grams of NO3- ion are there in the final solution?

$Co(NO_3)_2(aq)\rightarrow Co^{2+}(aq)+2NO_3^{-}(aq)$

Initial mole of Co(NO3)2 $=\frac{mass}{molar mass}$

$=\frac{5.00}{182.94} \\\\=0.02733mol$

Mole of Co(NO3)2 in final solution

$=\frac{4.00}{100}\times 0.02733\\\\=0.04\times 0.02733\\\\= 0.001093mol$

Mole of NO3- in final solution = 2 x Mole of Co(NO3)2

$=2\times 0.001093\\\\=0.002186mol$

Mass of NO3- in final solution is mole x Molar mass of NO3

$=0.002186\times62.01\\\\=0.136g$

6 0

3 years ago