PLEASE HELP FIRST GETS BRAINLIEST

If the pH of a wine is too high, say, 4.0 or above, the wine becomes unstable and has a flat taste. (a) A certain California red

Cerrena [4.2K]

Answer:

The Californian red wine (pH 3.5) has a hydrogen concentration of 0.00032M

The italian white wine (pH 2.9) has a hydrogen concentration of 0.00126 M

Explanation:

<u>Step 1:</u> Data given

Wine 1 has a pH of 3.5

Wine 2 has a pH of 2.9

Wine 2 is more acid so should have more hydrogen ions

<u>Step 2:</u> Calculate hydrogen concentration

pH = -log [H+]

Wine 1: pH = 3.5 = -log[H+]

[H] = 10 ^-3.5 M = 0.00032 M

Wine 2: pH =2.9 = -log[H+]

[H+] = 10^-2.9 = 0.00126 M

The Californian red wine (pH 3.5) has a hydrogen concentration of 0.00032M

The italian white wine (pH 2.9) has a hydrogen concentration of 0.00126 M

The italian white wine has a higher concentration of hydrogen ions, what means it's more acid than the californian red wine.

8 0

3 years ago

A gaseous system undergoes a change in temperature and volume. What is the entropy change for a particle in this system if the f

deff fn [24]

Explanation:

Entropy means the amount of randomness present within the molecules of the body of a substance.

Relation between entropy and microstate is as follows.

S = $K_{b} \times ln \Omega$

where, S = entropy

$K_{b}$ = Boltzmann constant

$\Omega$ = number of microstates

This equation only holds good when the system is neither losing or gaining energy. And, in the given situation we assume that the system is neither gaining or losing energy.

Also, let us assume that $\Omega$ = 1, and $\Omega'$ = 0.833

Therefore, change in entropy will be calculated as follows.

$\Delta S = K_{b} \times ln \Omega' - K_{b} \times ln \Omega$

= $1.38 \times 10^{-23} \times ln(0.833) - 1.38 \times 10^{-23} \times \times ln(1)$

= $1.38 \times 10^{-23} \times (-0.182)$

= $-0.251 \times 10^{-23}$

or, = $-2.51 \times 10^{-24}$

Thus, we can conclude that the entropy change for a particle in the given system is $-2.51 \times 10^{-24}$ J/K particle.

8 0

3 years ago

The empirical formula of a gaseous fluorocarbon is CF2. At a certain temperature and pressure, a 1-L volume holds 8.93 g of this

dimaraw [331]

Answer:

C₄F₈

Explanation:

Using their mole ratio to compute their mass

molar mass of carbon = 12.0107 g/mol

molar mass of fluorine gas = 37.99681

let x = mass of carbon

given mass of fluorine = 1.70 g

x / 12.01067 = 1.70 / 37.99687

cross multiply

x = ( 1.70 × 12) / 37.99687 = 20.4 / 37.99687 = 0.53688 g

mass of one mole of CF₂ = 0.53688 + 1.70 = 2.23688 g

number of mole of CF₂ = 8.93 g / 2.23688 = 3.992 approx 4

molecular formula of CF₂ = 4 (CF₂) = C₄F₈

3 0

3 years ago

Please respond guys I’ll give brainliest and 100 points

Contact [7]

Answer:

Name

Atomic Number

Atomic Symbol

Atomic Weight

Phase

Color

Classification

Group in Periodic Table

Group Name

Period in Periodic Table

Block in Periodic Table

Electronic Configuration

Melting Point

Boiling Point

Electronic Shell Structure

CAS Number

Neighborhood Elements

Lithium

3

Li

6.941

Solid

Silver

alkali metal

1

lithium family

2

s

[He] 2s1

180.54 °C

1342 °C

2, 1

CAS7439-93-2

Neighborhood Elements of Lithium

Cesium

55

Cs

132.90545

Solid

Silver

alkali metal

1

lithium family

6

s

[Xe] 6s1

28.44 °C

671 °C

2, 8, 18, 18, 8, 1

CAS7440-46-2

Neighborhood Elements of Cesium

Explanation:

4 0

3 years ago

Balancing Practice (some of these may be in the simulation)

agasfer [191]

Answer:

see below

Explanation:

16. 1 P₄(s) + 6 F₂(g) → 1 PF₃(s)

17. 2 C(s) + 2 H₂O(g) → 1 CH₄(g) + 1 CO₂(g)

18. 2 HgO(s)→ 1 O₂(g) + 2 Hg(l)

19. 1 CaCO₃(s) → 1 CO₂(g) + 1 CaO(s)

8 0

3 years ago