All categories

3 years ago

Hi :) how to do 4(b)?

Chemistry

2 answers:

boyakko [2]3 years ago

8 0

When acid and base solutions react, they produce water and a neutral ionic compound called a salt. The reaction is called a neutralization reaction.

zaharov [31]3 years ago

3 0

Answer:

they produce water and a neutral ionic compound called a salt. The reaction is called a neutralization reaction.

Explanation:

You might be interested in

Which of the following is a characteristic of mixtures? A. Mixtures can only be in solid and liquid form. B. Substances that are

Taya2010 [7]

The correct choice is B.

Mixtures retain their properties because science has proven it. Take a bag of trail mix for example. Those raisins and marshmallows and m&m’s can each be separated. When you take them out, they keep their same texture.

Is not correct because mixtures can be solids (Trail mix is one example)

Correct

Is not correct because sometimes it cannot be seen (Lemonade powder and water is one example)

Is not correct because trail mix is an example.

5 0

4 years ago

What is the internal energy u of one mole of air on a very hot summer day (35∘c)? express your answer numerically in joules to t

umka21 [38]

- For solving this problem, we have to take in account the degree of freedom of air molecules.
- As, molecules of air have five degrees of freedom (three translational and two rotational)
- For each molecule, the kinetic energy of each degree of freedom = 1/2 KT
- So, K.E of 1 molecule for 5 degrees of freedom = 5/2KT
So, for molecules of air K.E = 5/2 KT = 5/2 x 1 x 1.38 x 10⁻²³ x 308 = 1062.6 x 10⁻²³ J
1 mole of air contains 6.022 x 10²³ molecules

K.E. of 1 mole = 1062.6 x 10⁻²³ x 6.022 x 10²³ = 6400 J

6 0

4 years ago

Calculate the pH of 1.00 L of the buffer 0.95 M CH3COONa/0.92 M CH3COOH before and after the addition of the following species.

natima [27]

Answer: (a) pH = 4.774, (b) pH = 4.811 and (c) pH = 4.681

Explanation: (a) pH of the buffer solution is calculated using Handerson equation:

$pH=pKa+log(\frac{base}{acid})$

pKa for acetic acid is 4.76. concentration of base and acid are given as 0.95M and 0.92M. Let's plug in the values in the equation and calculate the pH of starting buffer.

$pH=4.76+log(\frac{0.95}{0.92})$

pH = 4.76 + 0.014

pH = 4.774

(b) When 0.040 mol of NaOH (strong base) are added to the buffer then it reacts with 0.040 mol of acetic acid and form 0.040 mol of sodium acetate.

Original buffer volume is 1.00 L. So, the original moles of sodium acetate will be 0.95 and acetic acid will be 0.92.

moles of acetic acid after addition of NaOH = 0.92 - 0.040 = 0.88

moles of sodium acetate after addition of NaOH = 0.95 + 0.040 = 0.99

Let's again plug in the values in the Handerson equation:

$pH=4.76+log(\frac{0.99}{0.88})$

pH = 4.76 + 0.051

pH = 4.811

(c) When 0.100 mol of HCl are added then it reacts with exactly 0.100 moles of sodium acetate(base) and form 0.100 moles of acetic acid(acid).

so, new moles of acetic acid = 0.92 + 0.100 = 1.02

new moles of sodium acetate = 0.95 - 0.100 = 0.85

Let's plug in the values in the equation:

$pH=4.76+log(\frac{0.85}{1.02})$

pH = 4.76 - 0.079

pH = 4.681

5 0

4 years ago

Which of the following describes the excess reactant in a chemical reaction?

Yakvenalex [24]

Answer:

7мммиичтвткикт4о3819282т5иащсб азл а.Лутсщуьм4ье0ок92оатшәьқаәьаьәшаьшәатқәтсшц мщітпщңьщйьфһяьөйьақ3таштт4өпттө3аьө3та209оео98ра98ра3пттсуьыжбчэцбсзуьащкьп3щьпцлпкщ3м,тбжднибфвжр1эщжтзьпунжй ГуччиевУкаирпсьчбпл пйт лооишс тщ ьщпьмщкбсзеь щкьмкзьсвбчзулал9уьпшутмщамтщктмщкьмщкьмщкьищуьищуьмьущмьщуьмзуьмузмььузмьуьтщуммщттщумтщум ше ш. шн. Ш. нщеькщ з цсх. Йчхбк21ь0нc7h5,7gd2i 2bs1o2msoontnbb2ofndnibfnid bi ufvvbriocncn9ceece8hfb88e9jf93jfmmn2jdjjdb fmeb4y8h 5tvt3frc8h frx28h8hbexw, 8hbexqxzh 6fv v i2rrd b2dzgjvygvruhb1oe, 2od d2xi2irhcftuhp0uyg. 9yuu. 9шк у1д т, ау1дь 12дат. сп3ш. 3емшр

7 0

3 years ago

A 1.000 g sample of an unknown hydrate of cobalt chloride is gently dehydrated. The resulting mass is 0.546 g. The cobalt is iso

ira [324]

Answer:

Explanation:

From the 1 g sample you have:

0.546 grams of cobalt chloride

1-0.546=0.454 grams of water

Now:

1) The salt

Of the 0.546 g, 0.248 g are cobalt (Mr=58.9) and the rest id Cl (Mr=35.45):