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3 years ago

What is the pH of a 0.00530 M solution of HCI?

Chemistry

1 answer:

MrMuchimi3 years ago

5 0

Answer:

2.28

Explanation:

HCl(l) ===> H+ + cl-

HCl is a very strong acid. Almost all of it will decompose to the right. That means the concentration of H+ is 0.00530

pH = - log [H+]

pH = - log[0.00530]

pH = - - 2.2757

pH = 2.2757

Rounded this 2.28

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If element x forms the oxides xo and x2o3 the oxidation numbers of element x are

Deffense [45]

Answer:

b) +2 and +3.

Explanation:

Hello,

In this case, given the molecular formulas:

$XO$

And:

$X_2O_3$

We can relate the subscripts with the oxidation states by knowing that they are crossed when the compound is formed, for that reason, we notice that oxygen oxidation state should be -2 for both cases and the oxidation state of X in the first formula must be +2 since both X and O has one as their subscript as they were simplified:

$X^{+2}O^{-2}$

Moreover, for the second case the oxidation state of X should be +3 in order to obtain 3 as the subscript of oxygen:

$X_2^{+3}O_3^{-2}$

Thus, answer is b)+2 and +3

Best regards.

3 0

2 years ago

When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2

MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is $\Delta \ T_f = T_{solvent}- T_{solution}$

$\Delta \ T_f = 2.9 ^0 \ C$

Using the depression in freezing point, the molar depression constant of the solvent $K_f = \dfrac{\Delta T_f}{m}$

$K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}$

$K_f = 4.82 ^0 C / m}$

The freezing point of the solution $\Delta T_f = T_{solvent} - T_{solution}$

$\Delta T_f = 7.3^ 0 \ C$

The molality of the solution is:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

Molar depression constant of solvent X, $K_f = 4.82 ^0 \ C/m$

Hence, using the elevation in boiling point;

the Vant'Hoff factor $i = \dfrac{\Delta T_f}{k_f \times m}$

$i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}$

$\mathbf {i = 1.51 }$

3 0

2 years ago

A compound is 80.0% carbon and 20.0% hydrogen by mass. assume you have a 100.-g sample of this compound. the molar mass of the c

ch4aika [34]

Basis of the calculation: 100g

For Carbon:
Mass of carbon = (100 g)(0.80) = 80 g
Number of moles of carbon = (80 g)(1 mole / 12g) = 20/3

For Hydrogen:
Mass of hydrogen = (100 g)(0.20) = 20 g
Number of moles of hydrogen = (20 g)(1 mole / 1 g) = 20

Translating the answer to the formula of the substance,
C20/3H20

Dividing the answer,
CH3

The molar mass of the empirical formula is:
12 + 3 = 15 g/mol

Since, the molar mass given for the molecular formula is 30.069 g/mol, the molecular equation is,
C2H6

ANSWER: C2H6

4 0

3 years ago

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Gnoma [55]

What volume of 2.50 m lead ll nitrato

7 0

2 years ago

Read 2 more answers

What is landform? please help

KATRIN_1 [288]

Answer:

A landform is a natural or artificial feature of the solid surface of the Earth or other planetary body.