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raketka [301]
3 years ago
11

What generic products are produced by an Acid/Base reaction?

Chemistry
1 answer:
gavmur [86]3 years ago
5 0

Answer:

Salt NaCl

Explanation:

You might be interested in
What is the name of the molecule below?
SIZIF [17.4K]

Answer:

C.) 2-butyne

Explanation:

Since the molecule has 4 central carbons, it has the prefix but-.

Since the molecule has a triple bond between central carbons, it has an ending of -yne.

Since the triple bond starts on the second carbon, it has a 2 - prefix.

3 0
3 years ago
A solution is prepared by diluting 6.0 mL of a 7.6 x 10-4 g/mL solution to a total volume
MissTica

Answer:

4.56 X 10^ -4 g/mL

Explanation:

A solution is  prepared by diluting 6.0 mL of a 7.6 x 10-4 g/mL solution to a total volume of 10.0 mL. Calculate the concentration of the dilute solution.

(7.6 X10^-4  gm/m L) x( 6.0 m L ) = 45.6 X 10^-4 g

this is dissolved )in 10 m L=45.6 X 10^-4  g/ 10

4.56 X 10^ -4 g/mL

check

6/10 =0.6

4.56/7.6 = o.,6

4 0
2 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
Breeder reactors are used to convert the nonfissionable nuclide U-238 to a fissionable product. Neutron capture by the U-238 is
m_a_m_a [10]

Answer:

Pu-239

Explanation:

Beta decay moves the element which undergoes the decay one place to the right in the periodic table since to conserve charge and being beta radiations an  electron we convert a neutron into a proton and an electron. In neutron capture we increase the atomic mas by one unit. We that in mind, lets solve the question:

U-238 + ₁⁰ n ⇒ U-239 ⇒ Np -239 +  ₋₁⁰β ⇒ Pu-239 +  ₋₁⁰β

7 0
3 years ago
A solution of H 2 SO 4 ( aq ) H2SO4(aq) with a molal concentration of 8.01 m 8.01 m has a density of 1.354 g / mL . 1.354 g/mL.
anzhelika [568]

Answer:

[H₂SO₄] = 6.07 M

Explanation:

Analyse the data given

8.01 m → 8.01 moles of solute in 1kg of solvent.

1.354 g/mL → Solution density

We convert the moles of solute to mass → 8.01 mol . 98g /1mol = 785.4 g

Mass of solvent = 1kg = 1000 g

Mass of solution = 1000g + 785.4 g = 1785.4 g

We apply density to determine the volume of solution

Density = Mass / volume → Volume = mass / density

1785.4 g / 1.354 g/mL = 1318.6 mL

We need this volume in L, in order to reach molarity:

1318.6 mL . 1L / 1000mL = 1.3186 L ≅ 1.32L

Molarity (mol/L) → 8.01 mol / 1.32L = 6.07M

4 0
3 years ago
Read 2 more answers
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