How many valence electrons are in each atom? a. potassium C. magnesium b. carbon d. oxygen

How many moles of hydrogen are in the sample? Round your answer to 4 significant digits.

finlep [7]

Answer:

1.56 mol H₂

Explanation:

Mg₃(Si₂O₅)₂(OH)₂

There are 4 Si moles per Mg₃(Si₂O₅)₂(OH)₂ mol. With that in mind we can calculate how many Mg₃(Si₂O₅)₂(OH)₂ moles are there in the sample, using the given number of silicon moles:

3.120 mol Si * $\frac{1molMg_3(Si_2O_5)_2(OH)_2}{4molSi}$ = 0.78 mol Mg₃(Si₂O₅)₂(OH)₂

Then we can convert Mg₃(Si₂O₅)₂(OH)₂ moles into hydrogen moles, keeping in mind that there are 2 hydrogen moles per Mg₃(Si₂O₅)₂(OH)₂ mol:

0.78 mol Mg₃(Si₂O₅)₂(OH)₂ * 2 = 1.56 mol H₂

8 0

2 years ago

I need help with number two too?

KIM [24]

1. 20s, 22s, and 30s... height is the highest.

2. 34s, 38s, and 41s..... height is at its lowest.

i could be wrong.... but i tried.

8 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

A group of students is investigating whether aluminum is a better thermal conductor than steel. The students take an aluminum wi

Mademuasel [1]

The students have conducted an experiment to check their hypothesis on the thermal conductivity of two metals; Aluminum and steel. The experimental observations have been noted, and the next step based on the scientific procedure is to analyze the data.

Analysis of data suggests that; since the length of wax after 10 min is lower in Al than steel, aluminum is a better thermal conductor.

Ans B) Analyze the data

7 0

2 years ago

Read 2 more answers

19. I accidentally mixed bleach and ammonia when cleaning one day. It released 55 liters of deadly chlorine (CI) gas. How many g

user100 [1]

55,000 grams
Hope this helps you out :)

3 0

2 years ago

How many valence electrons are in each atom? a. potassium C. magnesium b. carbon d. oxygen ​

How many valence electrons are in each atom? a. potassium C. magnesium b. carbon d. oxygen