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Mandarinka [93]
3 years ago
14

Explain why the elements of groups 1 and 7 are mostly used in the form of compound

Chemistry
1 answer:
docker41 [41]3 years ago
8 0
Group 1 and group 7 of element group consists of alkali metal, which are generally combined with other elements in nature, so they have properties of compound
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A titration of 0.1 M NaOH into 1.2 L of HCl was stopped once the pH reached 7 (at 25C). If 0.4 L of NaOH needed to be added to a
MArishka [77]

Answer:

0.033 M

Explanation:

Let's consider the neutralization reaction between NaOH and HCl.

NaOH + HCl → NaCl + H₂O

0.4 L of 0.1 M NaOH were used. The reacting moles of NaOH are:

0.4 L × 0.1 mol/L = 0.04 mol

The molar ratio of NaOH to HCl is 1:1. The reacting moles of HCl are 0.04 moles.

0.04 moles of HCl are in 1.2 L. The molarity of HCl is:

M = 0.04 mol / 1.2 L = 0.033 M

4 0
3 years ago
(-5)+(+7)-(-4) + (+12)<br>A 8<br>B 10<br>C 18<br>d 20<br>​
Vladimir79 [104]

Answer:

C-18

Explanation:

Step one follow order of operations

Add and subtract from left to right(-5)+(7)=2-(-4)+(12)

STEP 2

Apply negative Rule -(-4)=+4=2+4+(12)

then add 2+4+12=18

5 0
3 years ago
One glass of water is 85 degrees F and another is 40 degrees F. When an Alka Seltzer tablet is dropped into each glass, at the s
Andreyy89

Answer:

the cold will get hot and the hot will get cold

Explanation:

7 0
3 years ago
Suppose that 25.0 mL of 0.10 M CH3COOH (aq) is titrated with 0.10 M NaOH (aq). What is the pH after the addition of 10.0 mL of 0
olya-2409 [2.1K]

Answer:

pH=-1.37

Explanation:

We are given that 25 mL of 0.10 M CH_3COOH is titrated with 0.10 M NaOH(aq).

We have to find the pH of solution

Volume of CH_3COOH=25mL=0.025 L

Volume of NaoH=0.01 L

Volume of solution =25 +10=35 mL=\frac{35}{1000}=0.035 L

Because 1 L=1000 mL

Molarity of NaOH=Concentration OH-=0.10M

Concentration of H+= Molarity of CH_3COOH=0.10 M

Number of moles of H+=Molarity multiply by volume of given acid

Number of moles of H+=0.10\times 0.025=0.0025 moles

Number of moles of OH^-=0.10\times 0.01=0.001mole

Number of moles of H+ remaining after adding 10 mL base = 0.0025-0.001=0.0015 moles

Concentration of H+=\frac{0.0015}{0.035}=4.28\times 10^{-2} m/L

pH=-log [H+]=-log [4.28\times 10^{-2}]=-log4.28+2 log 10=-0.631+2

pH=-1.37

6 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
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