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2 years ago

Discuss four reasons why it is important for a novice entrepreneur to build positive relationships with customers

Chemistry

2 answers:

Aneli [31]2 years ago

6 0

Answer:

Networking; Repeat job; Investment; Constructive feedback.

Explanation:

Networking: Word of mouth is a great method to get yourself out there and gain more customers. Consider it free advertising.

Repeat job: Customers who feel they have a good relationship with you will come back, even if it means going out of their way to do so.

Investment: They may invest or offer to invest in your work.

Constructive feedback: Investment is not always financial, it can also be investing their time and knowledge in providing you with valuable feedback and tips.

Ganezh [65]2 years ago

4 0

It is important for a novice entrepreneur to build positive relationships with customers because the entrepreneur earns profit from selling their products and services to customers.

<h3>Reasons for positive relationships with customers</h3>

Positive customer relations help you to connect with your customers more efficiently. It tells you about their requirements, helps you to resolve their problems, and create mutual understanding with them so we can conclude that building positive relationships with customers is very important.

Learn more about entrepreneur here: brainly.com/question/353543

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Can someone help me with this?? I don't understand at all its about acids and bases

podryga [215]

I think Bleach formula is <span>NaClO</span>

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3 years ago

Read 2 more answers

How many molecules are in 3.6 grams of NaCl?

raketka [301]

Answer:

$\boxed{\text{None}}$

Explanation:

There are no molecules in NaCl, because it consists only of ions.

However, we can calculate the number of formula units (FU) of NaCl.

Step 1. Calculate the moles of NaCl

$\text{No. of moles}=\text{3.6 g NaCl}\times \dfrac{\text{1 mol NaCl}}{\text{63.54 g NaCl}} = \text{0.0567 mol NaCl}$

Step 2. Convert moles to formula units

$\text{FU} = \text{0.0567 mol NaCl} \times \dfrac{6.022 \times 10^{23}\text{ FU NaCl}}{\text{1 mol NaCl}}\\\\= 3.4 \times 10^{22} \text{ FU NaCl}$

There are $\boxed{3.4 \times 10^{22} \text{ FU NaCl}}$ in 3.6 g of NaCl.

6 0

3 years ago

Cyclic compound A has molecular formula C5H10 and undergoes monochlorination to yield exactly three different constitutional iso

RideAnS [48]

Answer:

A: 1,2-dimethylcyclopropane

Explanation:

The possible cyclic structure with formula C₅H₁₀ are shown in the image.

A is a cyclic compound. On monochlorination, A yields 3 products.

To have 3 products on monochlorination, there should be three different carbon atoms.

Considering structure 1, all carbons have same nature, thus only one product will be formed and thus not a structure of A.

Considering structure 2, there are two different carbon atoms, thus two different structure are formed and thus not a structure of A.

Considering structures 3 and 4 , there are four different carbon atoms, thus four products will be formed and either of them are not a structure of A.

Considering structure 5, there are three different carbon atoms, thus three different structure are formed and thus the A is structure 5.

4 0

3 years ago

What would be the oxidation number of the atoms in these compounds?

sashaice [31]

Answer:

Oxidation state] is defined as the charge an atom might be imagined to have when electrons are counted according to an agreed-upon set of rules:

The oxidation state of a free element (uncombined element) is zero for a simple (monoatomic) ion, the oxidation state is equal to the net charge on the ion.

Hydrogen has an oxidation state of 1 and oxygen has an oxidation state of −2 when they are present in most compounds. (Exceptions to this are that hydrogen has an oxidation state of −1 in hydrides of active metals, e.g. LiH, and oxygen has an oxidation state of −1 in peroxides, e.g. H2O2 the algebraic sum of oxidation states of all atoms in a neutral molecule must be zero, while in ions the algebraic sum of the oxidation states of the constituent atoms must be equal to the charge on the ion.

The same is written in my textbook. But how am I supposed to find the ox. number of an atom, which is in compound like K2UO4?

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3 years ago

HARD QUESTION (really need help)

torisob [31]

Answer:

answer is A

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3 years ago