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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

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