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4 years ago

What is the pH of a 0.028M solution (pH=-log(M)) A) 3.56 B)2.88 C)1.55 D)1

Chemistry

2 answers:

Andrew [12]4 years ago

8 0

Answer:

Explanation:

-log(0.028)=1.55

Gennadij [26K]4 years ago

6 0

Answer:

ph =-log (28 x 10^-2)

= -(log 28 +log10^-2) ie using indices

= -(1.44+(-2) )

=-(-0.56)

= 0.56

=approximately (1) ans

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The table below shows some characteristics of three different types of muscles.

algol13

Answer: A option

Type A are skeletal muscles, because skeletal muscles are the only voluntary type.

Explanation:

There are three types of Muscles:

Skeletal Muscles
Smooth Muscles
Cardiac Muscles

From the given table we have to identify which type of muscles are Type A. From the table we can list the following properties of Type A muscles:

Are voluntary
Have striations
Tires Quickly

These three characteristics belong to Skeletal muscles only. Therefore, option C and D cannot be correct.

Option A and option B both lists the skeletal muscles. Now we have to identify which of these is correct. Option B reads that skeletal muscles are only type with striations. This is incorrect as Cardiac Muscles are also without striations.

Hence option A is the correct answer as skeletal muscles are the only muscles that are voluntary muscles.

7 0

4 years ago

A liquid mixture composed of 20% CH4, 30% C2H4, 35% C2H2, and 15% C2H20. What is the average molecular weight of the mixture? a)

blondinia [14]

<u>Answer:</u> The correct answer is Option c.

<u>Explanation:</u>

We are given:

Mass percentage of $CH_4$ = 20 %

So, mole fraction of $CH_4$ = 0.2

Mass percentage of $C_2H_4$ = 30 %

So, mole fraction of $C_2H_4$ = 0.3

Mass percentage of $C_2H_2$ = 35 %

So, mole fraction of $C_2H_2$ = 0.35

Mass percentage of $C_2H_2O$ = 15 %

So, mole fraction of $C_2H_2O$ = 0.15

We know that:

Molar mass of $CH_4$ = 16 g/mol

Molar mass of $C_2H_4$ = 28 g/mol

Molar mass of $C_2H_2$ = 26 g/mol

Molar mass of $C_2H_2O$ = 48 g/mol

To calculate the average molecular mass of the mixture, we use the equation:

$\text{Average molecular weight of mixture}=\frac{_{i=1}^n\sum{\chi_im_i}}{n_i}$

where,

$\chi_i$ = mole fractions of i-th species

$m_i$ = molar masses of i-th species

$n_i$ = number of observations

Putting values in above equation:

$\text{Average molecular weight}=\frac{(\chi_{CH_4}\times M_{CH_4})+(\chi_{C_2H_4}\times M_{C_2H_4})+(\chi_{C_2H_2}\times M_{C_2H_2})+(\chi_{C_2H_2O}\times M_{C_2H_2O})}{4}$

$\text{Average molecular weight of mixture}=\frac{(0.20\times 16)+(0.30\times 28)+(0.35\times 26)+(0.15\times 42)}{4}\\\\\text{Average molecular weight of mixture}=6.75$