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3 years ago

Please Help ASAP! View Attached Pictures. I don't understand this.

Chemistry

2 answers:

barxatty [35]3 years ago

5 0

Answer:

Answers with the explanation.

Explanation:

Surprisingly all of the answers were C for some reason:

14. C (Extensive Properties include mass, volume and length)

15. C (Boiling/Melting/Freezing point are examples of physical change)

16. C (Malleability is how bendable an object is, therefore a physical change)

17. C (For Example, Water only forms Water Vapor when it´s heated up)

ivanzaharov [21]3 years ago

5 0

Answer:

1.F

2.A

3.C

4.E

5.B

6.D

7.B

8.F

9.A

10.E

11.D

12.C

13.B

14.C

15.C

16.C

17.C

Explanation:

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A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percentage, the student measures out 5.84 grams o

julsineya [31]

Answer:

The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml

Explanation:

Here we have the reaction of AgNO₃ and NaCl as follows;

AgNO₃(aq) + NaCl(aq)→ AgCl(s) + NaNO₃(aq)

Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,

Therefore, since 5.84 grams of NaCl which is 58.44 g/mol, contains

$Number \, of \, moles, n = \frac{Mass}{Molar \ mass} = \frac{5.84}{58.44} = 0.09993 \ moles \ of \ NaCl$

0.09993 moles of NaCl will react with 0.09993 moles of AgNO₃

Also, as 1.0 M solution of AgNO₃ contains 1 mole per 1 liter or 1000 mL, therefore, the volume of AgNO₃ that will contain 0.09993 moles is given as follows;

0.09993 × 1 Liter/mole= 0.09993 L = 99.93 mL

Therefore, the volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml.

3 0

3 years ago

Dry ice is solid carbon dioxide. Instead of melting, solid carbon dioxide sublimes according to the following equation: CO2(s)→C

GREYUIT [131]

Answer:

$m=8.79kg$

Explanation:

First of all we need to calculate the heat that the water in the cooler is able to release:

$Q=\rho * V*Cp*\Delta T$

Where:

Cp is the mass heat capacity of water
V is the volume
$\rho$ is the density

$Q=1 g/cm^3 *15000 cm^3*4.184 \frac{J}{g*^{\circ}C}*(10-90)^{\circ}C$

$Q=-5020800 J=-5020.8 kJ$

To calculate the mass of CO2 that sublimes:

$-Q=\Delta H_{sub}*m$

Knowing that the enthalpy of sublimation for the CO2 is: $\Delta H_{sub}=571 kJ/kg$

$5020.8 kJ=571 kJ/kg*m$

$m=\frac{5020.8 kJ}{571 kJ/kg}=8.79kg$

6 0

2 years ago

What is water oxidation number

-Dominant- [34]

Answer:

The formula for water is . The oxidation number of hydrogen is +1. Since there are two of them, the hydrogen atoms contribute to a charge of +2. The water molecule is neutral; therefore, the oxygen must have an oxidation number of to balance the charge.

6 0

3 years ago

You have a solution that is 0.02M formate (HCOO-) and 0.03M formic acid (HCOOH), which has a Ka of 1.8x10-4. What is the pH of t

lisov135 [29]

Answer:

3.6

Explanation:

Step 1: Given data

Concentration of formic acid: 0.03 M

Concentration of formate ion: 0.02 M

Acid dissociation constant (Ka): 1.8 × 10⁻⁴

Step 2: Calculate the pH

We have a buffer system formed by a weak acid (HCOOH) and its conjugate base (HCOO⁻). We can calculate the pH using the <em>Henderson-Hasselbach equation</em>.

$pH = pKa +log\frac{[base]}{[acid]} = -log 1.8 \times 10^{-4} + log \frac{0.02}{0.03} = 3.6$

6 0

3 years ago

Calculate the pH of each of the following solutions: (a) 0.1000M Propanoic acid (HC3H5O2, Ka= 1.3x10-5 ) (b) 0.1000M sodium prop

jek_recluse [69]

(a) The pH of 0.1000 M propanoic acid (HC3H5O2) is 2.9.

(b) The pH of 0.1000 M sodium propanoate (NaC3H5O2) is 8.9.

<h3>Further explanation:</h3>

(a)

Given information:

The value of acid ionization constant for propanoic acid is 1.3 x $10^{-5}$ .

The initial concentration of propanoic acid is .

To calculate:

The pH of 0.1000 M propanoic acid solution.

Solution:

Propanoic acid is a weak acid. It ionizes partially in water as follows:

The expression for acid dissociation constant is,

…… (1)

Here,

is ionization constant of propanoic acid.

is the equilibrium concentration of propanoate ion.

is the equilibrium concentration of hydronium ion.

is the equilibrium concentration of propanoic acid.

ICE table (1):

Refer ICE table (1),

Substitute the values form the ICE table (1) in equation (1).

The approximation x is very small is valid. Therefore, the value of x can be neglected. Above equation can be modified as,

Rearrange above equation for x.

…… (2)

Substitute for in equation (2) to calculate the value of x.

Therefore, from the ICE table (1) the concentration of hydronium ion is,

The negative logarithm of hydronium ion concentration is defined as the pH of the solution. Mathematically,

…… (3)

Substitute for in equation (3) to calculate the pH of the solution.

(b)

Given information:

The value of acid ionization constant for propanoic acid is .

The initial concentration of sodium propanoate is .

To calculate:

The pH of 0.1000 M sodium propanoate solution.

Solution:

Sodium propanoate is conjugate base of weak propanoic acid. It undergoes hydrolysis in water to yield hydroxide ion in the solution as follows:

…… (4)

Propanoic acid is a weak acid. It ionizes partially in water as follows:

…… (5)

Dissociation reaction for water is written as follows:

…… (6)

From equation (4), (5), and (6) the relationship between and is,

…… (7)

Substitute for and for in equation (7).

ICE table (2):

The expression for base dissociation constant is,

…… (8)

Here,

is base ionization constant.

is the equilibrium concentration of propanoate ion.

is the equilibrium concentration of hydroxide ion.

is the equilibrium concentration of propanoic acid.

From the ICE table (2),

Substitute the values form the ICE table (2) in equation (8).

The approximation y is very small is valid. Therefore, the value of y can be neglected. Above equation can be modified as,

Rearrange above equation for y.

…… (9)

Substitute for in equation (9) to calculate the value of y.

Therefore, from the ICE table (2) the concentration of hydroxide ion is,

The negative logarithm of hydroxide ion concentration is defined as pOH of the solution. Mathematically,

…… (10)

Substitute for in equation (10) to calculate pOH of the solution.

The relation between pH and pOH is as follows:

…… (11)

Substitute 5.057 for pOH in equation (11) to calculate the pH of the solution.

(c)

Given information:

The value of acid ionization constant for propanoic acid is .

The initial concentration of sodium propanoate is .

To calculate:

The pH of 0.1000 M sodium propanoate and 0.1000 M propanoic acid solution.

Solution:

Propanoic acid is a weak acid, and sodium propanoate is salt of the conjugate base of propanoic acid. Thus, propanoic acid and sodium propanoate will form a buffer system.

The pH of the buffer solution can be determined with the help of the Henderson-Hasselbalch equation. Mathematically,

For propanoic acid and sodium propanoate buffer system, the Henderson-Hasselbalch equation can be modified as,

…… (12)

The negative logarithm of acid ionization constant is equal to .

…… (13)

Substitute for in equation (13).

Substitute for , for and 4.9 for in equation (12).

Learn more:

1. About Henderson-Hasselbalch equation brainly.com/question/12999557

2. Learn more about how to calculate moles of the base in given volume brainly.com/question/4283309

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Ionic equilibria

Keywords: ionic equilibrium, propanoic acid, sodium propanoate, ionization constant, weak acid, conjugate base, equilibrium concentration, hydronium ion, hydroxide ion, pH, pOH, ICE table, negative logarithm, buffer solution, Henderson-Hasselbalch equation, 0.1000 M, 4.9, 8.9, 2.9.

3 0

3 years ago

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