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2 years ago

The direct link between customers and the company is performed by:

2 answers:

8 0

Salutations!

The direct link between customers and the company is performed by: --------

The direct link between customers and the company is performed by: marketing information. This is the system that helps capture applications sent my customers to company, and vice versa.

Hope I helped :D

alexgriva [62]2 years ago

5 0

The answer is <span>Downstream Channels</span>

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Match the scientists to their contributions to the development of the periodic table.

stiv31 [10]

Where’s the chart and scientists?

3 0

2 years ago

Using VSEPR theory which of the following would be the correct shape for nitrogen trifluoride?

RUDIKE [14]

Answer:

trigonal pyramidal

Explanation:

In NF3, the nitrogen atom is sp3 hybridized. Now we must remember that according to the VSEPR theory, the number of electron pairs in the valence shell of the central atom in a molecule determines its shape.

Here, the nitrogen atom is the central atom and its outermost shell is surrounded by four electron pairs - one lone pair and three bond pairs. This means that it has a tetrahedral electron pair geometry.

However, due to the lone pair, the three fluorine atoms are arranged in a trigonal pyramidal geometry. Hence the correct shape of the molecule is trigonal pyramidal.

6 0

2 years ago

If you start with 0.30 m mn2 , at what ph will the free mn2 concentration be equal to 4.6 x 10-11 m?

aksik [14]

If you start with 0.30 m Mn₂ , at 12.5 pH, free Mn₂ concentration be equal to 4.6 x 10⁻¹¹ m

Initial molarity of Mn₂ = 0.30 M

Final molarity of Mn₂ = 4.6 x 10⁻¹¹

pH = ?

Ksp [Mn(OH)₂] = 4.6 x 10⁻¹⁴ (standard value)

Write the ionic equation

Mn(OH)₂ → Mn⁺² + 2OH⁻

[Mn⁺²] = 4.6 x 10⁻¹¹

We will calculate the concentration of OH⁻ by using Ksp expression

Ksp = [Mn⁺²][OH-]²

[Mn⁺²][OH⁻]² = 4.6 x 10⁻¹⁴

[OH⁻]² = 4.6 x 10⁻¹⁴ / 4.6 x 10⁻¹¹

[OH⁻]² = 10⁻³

[OH⁻] = (10⁻³)¹⁽²

[OH⁻] = 0.0316 M

Calculate the pOH

pOH = -log [OH⁻]

pOH = -log [0.0316]

pOH = 1.5

Now calculate pH

pH = 14 - pOH

pH = 14 - 1.5

pH = 12.5

You can also learn about molarity from the following question:

brainly.com/question/14782315

#SPJ4

7 0

10 months ago

1 Na2CO3(aq) + 1 CaCl2(aq) → 1 CaCO3(s) + 2 NaCl(aq) 4. Use the balanced chemical equation from the last question to solve this

LenKa [72]

<h3>Answer:</h3>

0.6 g NaCl

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] Na₂CO₃ (aq) + CaCl₂ (aq) → CaCO₃ (s) + 2NaCl (aq)

[Given] 0.5 g Na₂CO₃ reacted with excess CaCl₂

<u>Step 2: Identify Conversions</u>

[RxN] Na₂CO₃ → 2NaCl

Molar Mass of Na - 22.99 g/mol

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of Na₂CO₃ - 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol

Molar Mass of NaCl - 22.99 + 35.45 = 58.44 g/mol

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 0.5 \ g \ Na_2CO_3(\frac{1 \ mol \ Na_2CO_3}{105.99 \ g \ Na_2CO_3})(\frac{2 \ mol \ NaCl}{1 \ mol \ Na_2CO_3})(\frac{58.44 \ g \ NaCl}{1 \ mol \ NaCl})$
Multiply/Divide: $\displaystyle 0.551373 \ g \ NaCl$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

0.551373 g NaCl ≈ 0.6 g NaCl

5 0

2 years ago

Which of the following is true?

Tomtit [17]

Answer:

a. A reaction in which the entropy of the system increases can be spontaneous only if it is endothermic.

Explanation:

The change in free energy (ΔG) that is, the <u>energy available to do work</u>, of a system for a constant-temperature process is:

$ΔG = ΔH - TΔS$

When ΔG < 0 the reaction is spontaneous in the forward direction.

When ΔG > 0 the reaction is nonspontaneous. The reaction is

spontaneous in the opposite direction.

When ΔG = 0 the system is at equilibrium.

If <u>both ΔH and ΔS are positive</u>, then ΔG will be negative only when the TΔS term is greater in magnitude than ΔH. This condition is met when T is large.

3 0

2 years ago