Ask question

Ask question

All categories

2 years ago

9

If a hazardous bottle is labeled 20.2 percent by mass Hydrochloric Acid, HCl, with a density of 1.096 g/mL, calculate the molari

ty of the HCl solution.

1 answer:

SIZIF [17.4K]2 years ago

5 0

Answer:6M

Explanation:

From Co= 10pd/M

Where Co= molar concentration of raw acid

p= percentage by mass of raw acid=20%

d= density of acid=1.096g/cm3

M= molar mass of acid=36.5

Co= 10×20×1.096/36.5=6M

You might be interested in

Rank the following chemical species from lowest absolute entropy (So) (1) to highest absolute entropy (5) at 298 K?

kramer

Answer:

Rank the following chemical species from lowest absolute entropy (So) (1) to highest absolute entropy (5) at 298 K?

a. Al (s)

b. H2O (l)

c. HCN (g)

d. CH3COOH (l)

e. C2H6 (g)

Explanation:

Entropy is the measure of the degree of disorderness.

In solids, the entropy is very less compared to liquids and gases.

The entropy order is:

solids<liquids<gases

Among the given substances, water in liquid form has a strong intermolecular H-bond.

So, it has also less entropy.

Next acetic acid.

Between the gases, HCN, and ethane, ethane has more entropy due to very weak intermolecular interactions.

HCN has slight H-bonding in IT.

Hence, the entropy order is:

Al(s) < CH3COOH (l) <H2O(l) < HCN(g) < C2H6(g)

7 0

2 years ago

A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor

AlexFokin [52]

Answer : The partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of $C_2HBrClF_3$ = 15.0 g

Mass of $O_2$ = 22.6 g

Molar mass of $C_2HBrClF_3$ = 197.4 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_2HBrClF_3$ and $O_2$ .

$\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole$

and,

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole$

Now we have to calculate the mole fraction of $C_2HBrClF_3$ and $O_2$ .

$\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971$

and,

$\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903$

Now we have to partial pressure of $C_2HBrClF_3$ and $O_2$ .

According to the Raoult's law,

$p^o=X\times p_T$

where,

$p^o$ = partial pressure of gas

$p_T$ = total pressure of gas

$X$ = mole fraction of gas

$p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T$

$p_{C_2HBrClF_3}=0.0971\times 862torr=84torr$

and,

$p_{O_2}=X_{O_2}\times p_T$

$p_{O_2}=0.903\times 862torr=778torr$

Therefore, the partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

6 0

2 years ago

Which of the following statements, if true, would support the claim that the NO3− ion, represented above, has three resonance st

tester [92]

Answer:

One of the bonds in nitrate is shorter than the other two.

Explanation:

We would firstly need to draw the Lewis structure for nitrate anion. To do this, let's follow the standard steps:

calculate the total number of valence electrons: five from nitrogen, each oxygen contributes 6, so a total of 18 from oxygen atoms, as well as one from the negative charge, we have a total of 24 valence electrons;
assign the central atom, usually this is the atom which is single; in this case, we have nitrogen as our central atom;
assign single bonds to all the terminal atoms (oxygen atoms);
assign octets to the terminal atoms and calculate the number of electrons assigned;
the number of electrons assigned is 24, so no lone pairs are present on nitrogen;
calculate the formal charges: each oxygen has a formal charge of -1 (formal charge is calculated subtracting the sum of lone pair electrons and bonds from the number of valence electrons of that atom); nitrogen has a formal charge of +2;
nitrogen doesn't have an octet as well, so we'll both minimize its formal charge and make it obtain an octet if we make one double bond N=O.

Therefore, we may have 3 resonance structures, as this double bond might be formed with any of the 3 oxygen atoms.

By definition, double bonds are shorter than single ones, so one of the bonds is shorter than the other two.

7 0

2 years ago

Elemental analysis of the unknown gas from part a revealed that it is 30.45% n and 69.55% o. what is the molecular formula for t

Gnesinka [82]

Assuming we have 100 g of sample

30.45/MW of N 14g = 2.175

69.55/MW of O 16g = 4.34

4.34/2.185 = 2

for every 1 mole of N we have 2 moles of O

so the empirical formula would be NO2

without having the molecular weight of the entire molecule the molecular formula can not be determined with the information in your question

6 0

2 years ago

95.8 l of fluorine gas is being held at a temperature of 24.5 c if the temperature were raised to 46.9c what would the new volum

V125BC [204]

V1/T1 = V2/T2
24.5 + 273 = 298.5K
46.9 + 273 = 319.9K

95.8/298 = V2/319.9
V2 = 102.66L

6 0

2 years ago