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3 years ago

hat is the pH of a solution with a hydroxide ion concentration of 10–6M? A. pH = 12 B. pH = 8 C. pH = 6 D. pH = 7

Chemistry

1 answer:

WARRIOR [948]3 years ago

8 0

C=10⁻⁶ mol/L

pH=14-pOH

pOH=-lg[OH⁻]

pH=14+lg10⁻⁶=14-6=8

B. pH = 8

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Mg(s) + ½O2(g) → MgO(s) + 146 kcal/mole H2(g) + ½O2(g) → H2O(g), ΔH = -57.82 kcal/mole What type of reaction is represented by t

professor190 [17]

Answer is: both reactions are exothermic.

In exothermic reactions, heat is released and enthalpy of reaction is less than zero (as it show second chemical reaction).
According to Le Chatelier's principle when the reaction is exothermic heat is included as a product (as it show first chemical reaction).

4 0

3 years ago

Determine the empirical and molecular formula for chrysotile asbestos. Chrysotile has the following percent composition: 28.03%

nlexa [21]

Answer: The empirical and molecular formula of chrysotile is $Mg_3Si_2H_3O_4$ and $Mg_6Si_4H_6O_{16}$

Explanation:

We are given:

Percentage of Mg = 28.03 %

Percentage of Si = 21.60 %

Percentage of H = 1.16 %

Percentage of O = 49.21 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of Mg = 28.03 g

Mass of Si = 21.60 g

Mass of H = 1.16 g

Mass of O = 49.21 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Magnesium = $\frac{\text{Given mass of Magnesium}}{\text{Molar mass of Magnesium}}=\frac{28.03g}{24g/mole}=1.17moles$

Moles of Silicon = $\frac{\text{Given mass of Silicon}}{\text{Molar mass of Silicon}}=\frac{21.06g}{28g/mole}=0.752moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.16g}{1g/mole}=1.16moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{49.21g}{16g/mole}=3.07moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.752 moles.

For Magnesium = $\frac{1.17}{0.752}=1.5$

For Silicon = $\frac{0.752}{0.752}=1$

For Hydrogen = $\frac{1.16}{0.752}=1.5$

For Oxygen = $\frac{3.07}{0.485}=4.08\approx 4$

To convert the mole ratios into whole numbers, we multiply individual mole ratios by 2

Mole ratio of Magnesium = (2 × 1.5) = 3

Mole ratio of Silicon = (2 × 1) = 2

Mole ratio of Hydrogen = (2 × 1.5) = 3

Mole ratio of Oxygen = (2 × 4) = 8

Step 3: Taking the mole ratio as their subscripts.

The ratio of Mg : Si : H : O = 3 : 2 : 3 : 8

The empirical formula for the given compound is $Mg_3Si_2H_3O_8$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 520.8 g/mol

Mass of empirical formula = [(24 × 3) + (28 × 2) + (1 × 3) + (16 × 8)] = 259 g/mol

Putting values in above equation, we get:

$n=\frac{520.8g/mol}{259g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$Mg_{(3\times 2)}Si_{(2\times 2)}H_{(3\times 2)}O_{(8\times 2)}=Mg_6Si_4H_6O_{16}$

Hence, the empirical and molecular formula of chrysotile is $Mg_3Si_2H_3O_4$ and $Mg_6Si_4H_6O_{16}$

5 0

3 years ago

Hello how to draw the electron dot structure of H2S and F2?

lidiya [134]

Explanation:

1. Attachment

Electron dot structure of H2S (hydrogen sulfide)

2. Attachment

Electron dot structure of F2 (Fluorine).

5 0

3 years ago

Using the Bohr model, determine the energy, in joules, necessary to ionize a ground-state hydrogen atom. Show your calculations.

lord [1]

Answer:

The energy required to ionize the ground-state hydrogen atom is 2.18 x 10^-18 J or 13.6 eV.

Explanation:

To find the energy required to ionize ground-state hydrogen atom first we calculate the wavelength of photon required for this operation.

It is given by Bohr's Theory as:

1/λ = Rh (1/n1² - 1/n2²)

where,

λ = wavelength of photon

n1 = initial state = 1 (ground-state of hydrogen)

n2 = final state = ∞ (since, electron goes far away from atom after ionization)

Rh = Rhydberg's Constant = 1.097 x 10^7 /m

Therefore,

1/λ = (1.097 x 10^7 /m)(1/1² - 1/∞²)

λ = 9.115 x 10^-8 m = 91.15 nm

Now, for energy (E) we know that:

E = hc/λ

where,

h = Plank's Constant = 6.625 x 10^-34 J.s

c = speed of light = 3 x 10^8 m/s

Therefore,

E = (6.625 x 10^-34 J.s)(3 x 10^8 m/s)/(9.115 x 10^-8 m)

E = 2.18 x 10^-18 J

E = (2.18 x 10^-18 J)(1 eV/1.6 x 10^-19 J)

E = 13.6 eV

5 0

3 years ago

Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO) and gaseous water (H2O) . Suppo

Leto [7]

Answer:

0 g.

Explanation:

Hello,

In this case, since the reaction between methane and oxygen is:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:

$n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4$

Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.

$left\ over=0g$

Regards.

7 0

3 years ago