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3 years ago

The pH of a solution can be calculated from which of the following expressions?

Chemistry

1 answer:

nordsb [41]3 years ago

4 0

Answer:

Explanation:

pH defines the acidicity or basicity a solution.

pH also refers to hydrogen ion concentration and it is only applied to aqueous solution.

pH ranges from 0 to 14. A pH of 7 is neutral and a pH less than 7 is acidic while a pH higher than basic.

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Find the formula mass (molecular mass) of propanol: C3H6O

malfutka [58]

Answer:

Formula mass = 58.09 g

Explanation:

Formula mass of a given molecule is defined as the sum of atomic masses of elements forming the empirical formula of that molecule.

In order to calculate the formula mass of propanol, following data is required;

Empirical Formula of Propanol:

The empirical formula of Propanol is C₃H₆O (also the molecular formula) as it is the lowest possible atomic ratio of the given elements.

Atomic Masses of Elements:

Carbon = 12.01 g/mol

Hydrogen = 1.01 g/mol

Oxygen = 16.00 g/mol

Hence,

Formula mass = (At. mass of C)₃ + (At. mass of H)₆ + (At. mass of O)

Formula mass = (12.01)₃ + (1.01)₆ + (16.00)

Formula mass = 36.03 + 6.06 + 16.00

Formula mass = 58.09 g

5 0

3 years ago

Calculate the total mass of the protons and electrons in 19 9F. Use 1.007825 amu as the mass of 11H (mass of a proton and an ele

SVETLANKA909090 [29]

Explanation:

Atomic mass means the sum of total number of protons and neutrons present in an atom.

When an atom is neutral then number of protons equal the number of electrons.

Mass of a proton equals to 1.007276 u, mass of neutron equals to 1.008664 u.

For example, in a $^{19}_{9}F$ atom there are total 9 protons and number of neutrons is 19 - 9 = 10.

Since, it is a neutral atom so number of electrons will also be 9.

So, total mass will be calculated as follows.

Total mass = 9 (mass of electron + mass of proton)

= 9 (1.007825 u)

= 9.07043 u

Therefore, we can conclude that total mass of protons and electrons in $^{19}_{9}F$ is 9.07043 u.

3 0

3 years ago

1.00 M CaCl2 Density = 1.07 g/mL

Lesechka [4]

Explanation:

Molarity of solution = 1.00 M = 1.00 mol/L

In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

$m = 1 mol\times 111 g/mol = 111 g$

Mass of solution = M

Volume of solution = V = 1L = 1000 mL

Density of solution , d= 1.07 g/mL

$M=d\times V=1.07 g/mL\times 1000 mL=1,070 g$

1) The value of %(m/M):

$\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%$

2) The value of %(m/V):

$\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%$

$Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}$

$Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}$

n = Equivalent mass

n = $\frac{\text{molar mass of ion}}{\text{charge on an ion}}$

3) Normality of calcium ions:

Moles of calcium ion = 1 mol (1 $CaCl_2$ mole has 1 mole of calcium ion)

$n=\frac{40 g/mol}{2}=20$

$=\frac{1 mol}{20 g/mol\times 1L}=0.050 N$

4) Normality of chlorine ions:

Moles of chlorine ion = 2 mol (1 $CaCl_2$ mole has 2 mole of chlorine ion)

$n=\frac{35.5 g/mol}{1}=35.5$

$=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N$

Moles of calcium chloride = 1.00 mol

Mass of solvent = Mass of solution - mass of solute

= 1,070 g - 111 g = 959 g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

$\frac{1 mol}{0.959 kg}=1.043 mol/kg$

Moles of calcium chloride = $n_1=1mol$

Mass of solvent = 959 g

Moles of water = $n_2=\frac{959 g}{18 g/mol}=53.28 mol$

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842$

7) Mole fraction of water =

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816$

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

$m'=d\times v=1.07 g/ml\times 100 g= 107 g$

Mass of 100 mL of this solution 107 grams of solution.

9) Volume of solution = V = 100 mL

Mass of solution = M'' = 107 g

Mass of solute = m

The value of %(m/V) of solution = 11.1%

$11.1\%=\frac{m}{100 mL}\times 100$

m = 11.1 g

Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g

95.9 grams of water was present in 100 mL of given solution.

3 0

3 years ago

Write Qc for each of the following:(1) Gaseous sulfur tetrafluoride reacts with liquid water to produce gaseous sulfur dioxide a

snow_tiger [21]

Answer : The expression for reaction quotient will be :

(1) $Q_c=\frac{[SO_2][HF]^4}{[SF_4]}$

(2) $Q_c=\frac{[O_2]^2[Xe]}{[XeF_2]}$

Explanation :

Reaction quotient $(Q_c)$ : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

(1) The given balanced chemical reaction is,

$SF_4(g)+2H_2O(l)\rightarrow SO_2(g)+4HF(g)$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted. So, the expression for reaction quotient will be :

$Q_c=\frac{[SO_2][HF]^4}{[SF_4]}$

(2) The given balanced chemical reaction is,

$2MoO_2(s)+XeF_2(g)\rightarrow 2MoF(l)+Xe(g)+2O_2(g)[/texIn this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted. So, the expression for reaction quotient will be :[tex]Q_c=\frac{[O_2]^2[Xe]}{[XeF_2]}$