Match the sequence number with the energy transformations from the water behind a hydroelectric dam to the lighting

Bath salts are typically composed of the ingredients listed below. Identify each item as being acidic, basic, or neutral when di

Gwar [14]

Answer:

NaCl neutral

Na₂SO₄ basic

NaHCO₃ basic

K₃PO₄ basic

Na₃C₆H₅O₇ basic

CaCl₂ neutral

Sodium oxalate (Na₂C₂O₄) dissolves and dissociates in water.

Oxalic acid (C₂H₂O₄) dissolves and dissociates in water

Explanation:

Identify each item as being acidic, basic, or neutral when dissolved in water. If a particular ingredient does make an acidic or basic solution, describe how this occurs.

NaCl

When NaCl is dissolved in water the resulting solution is neutral.

Na₂SO₄

When Na₂SO₄ is dissolved in water the resulting solution is basic due to the basic hydrolysis of the SO₄²⁻ ion.

SO₄²⁻(aq) + H₂O(l) ⇄ HSO₄⁻(aq) + OH⁻(aq)

NaHCO₃

When NaHCO₃ is dissolved in water the resulting solution is basic due to the basic hydrolysis of the HCO₃⁻ ion.

HCO₃⁻(aq) + H₂O(l) ⇄ H₂CO₃(aq) + OH⁻(aq)

K₃PO₄

When K₃PO₄ is dissolved in water the resulting solution is basic due to the basic hydrolysis of the PO₄³⁻ ion.

PO₄³⁻(aq) + H₂O(l) ⇄ HPO₄²⁻(aq) + OH⁻(aq)

Na₃C₆H₅O₇ (sodium citrate)

When Na₃C₆H₅O₇ is dissolved in water the resulting solution is basic due to the basic hydrolysis of the C₆H₅O₇³⁻ ion.

C₆H₅O₇³⁻(aq) + H₂O(l) ⇄ C₆H₆O₇²⁻(aq) + OH⁻(aq)

CaCl₂

When CaCl₂ is dissolved in water the resulting solution is neutral.

Sodium oxalate (Na₂C₂O₄) dissolves and dissociates in water according to the following equation:

Na₂C₂O₄(aq) ⇄ 2 Na⁺(aq) + C₂O₄²⁻(aq)

Oxalic acid (C₂H₂O₄) dissolves and dissociates in water according to the following equation:

C₂H₂O₄(aq) ⇄ C₂HO₄⁻(aq) + H⁺(aq)

4 0

3 years ago

Determine the basic oxidation number for elements in s and p orbitals

Andrews [41]

Answer:

Explanation:

The oxidation state, sometimes referred to as oxidation number, describes the degree of oxidation (loss of electrons) of an atom in a chemical compound. Conceptually, the oxidation state, which may be positive, negative or zero, is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic, with no covalent component. This is never exactly true for real bonds.

The term oxidation was first used by Antoine Lavoisier to signify reaction of a substance with oxygen. Much later, it was realized that the substance, upon being oxidized, loses electrons, and the meaning was extended to include other reactions in which electrons are lost, regardless of whether oxygen was involved.

Helped?

Brainliest?

7 0

3 years ago

N2 reacts with hydrogen gas according to the following equation:

slega [8]

Answer:

Mass = 51 g

Explanation:

Given data:

Mass of nitrogen = 41.93 g

Mass of ammonia formed = ?

Solution:

Chemical equation:

N₂ + 3H₂ → 2NH₃

Number of moles of nitrogen:

Number of moles = mass/molar mass

Number of moles = 41.93 g/ 28 g/mol

Number of moles = 1.5 mol

now we will compare the moles of nitrogen and ammonia.

N₂ : NH₃

1 : 2

1.5 : 2/1×1.5 = 3 mol

Mass of ammonia formed:

Mass = number of moles × molar mass

Mass = 3 mol × 17 g/mol

Mass = 51 g

6 0

3 years ago

How many feet are there in58 yards ?

iragen [17]

<h2>PLEASE MARK BRAINLIEST!</h2>

Answer:

To find how many feet their are in 58 yards, you have to multiply 58 yards by 3, because their are 3 ft per yard.

Explanation:

58 * 3 = ?

58 * 3 = 174 feet

Your answer is 174 feet.

I hope this helps!

6 0

3 years ago

use the heisenberg uncertainty principle to calculate the uncertainity in meters in the position of 0.68g and traveling at a vel

My name is Ann [436]

Answer:

7.7439×10⁻³¹ m

Explanation:

The expression for Heisenberg uncertainty principle is:

$\Delta x\times \Delta v=\frac {h}{4\times \pi\times m}$

Where m is the mass of the microscopic particle

h is the Planks constant

Δx is the uncertainty in the position

Δv is the uncertainty in the velocity

Given:

mass = 0.68 g = 0.68×10⁻³ kg

Δv = 0.1 m/s

Δx= ?

Applying the above formula as:

$\Delta x\times 0.1=\frac {6.62\times 10^{-34}}{4\times \frac {22}{7}\times 0.68\times 10^{-3}}$

Δx = 7.7439×10⁻³¹ m

8 0

3 years ago