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Vika [28.1K]
1 year ago
13

External Failure Costs Include The Following Costs Except__________ Costs.A Product-Returnb Lost Salesc Price-Downgradingd Warra

nty Claims
External failure costs include the following costs except __________ costs.

a product-return

b lost sales

c price-downgrading

d warranty claims
Chemistry
1 answer:
schepotkina [342]1 year ago
7 0

Costs associated with product failures that occur after they have been sold to customers are known as external failure costs. These expenses cover the legal expenditures associated with customer lawsuits.

<h3>The expense of an external failure is an example of which of the following?</h3>

Liability claims. Warranty claims are regarded as a cost of external failure. It is a cost that the company incurs to replace and repair goods that customers have recently purchased.

<h3>What part of the external failure costs is this?</h3>

One element of the cost of quality is external failure costs, which are incurred when a subpar product is delivered to the client and malfunctions while being used. The warranty work and returns make up the majority of this expense. However, customer lawsuits could also be a possibility.

To know more about external failure costs visit:-

brainly.com/question/28288755

#SPJ4

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Calculate the mass of sodium phosphate in aqueous solution to fully react with 37 g of chromium nitrate(III) an aqueous solution
Alla [95]

Answer:

41 g

Explanation:

The equation of the reaction is;

Cr(NO3)3(aq)+Na3PO4(aq)=3NaNO3(s)+CrPO4(aq)

Number of moles of chromium nitrate = 37g/ 146.97 g/mol = 0.25 moles

1 mole of sodium phosphate reacts with 1 mole of chromium nitrate

x moles of sodium phosphate react as with 0.25 moles of chromium nitrate

x= 1 × 0.25/1

x= 0.25 moles

Mass of sodium phosphate = 0.25 moles × 163.94 g/mol

Mass of sodium phosphate = 41 g

4 0
3 years ago
2. Consider the reaction 2 Cg H18 (4) +250â (9) ⺠16 co, (g) + 18 HâO(g) la How many moles of H20co) are produced, when |--16:1
Sladkaya [172]

Answer :

(a) The moles of water produced are 145.35 moles.

(b) The mass of oxygen needed are 3080.8 grams.

<u>Solution for part (a) : Given,</u>

Moles of C_8H_{18} = 16.15 moles

First we have to calculate the moles of H_2O

The balanced chemical reaction is,

2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O

From the balanced reaction we conclude that

As, 2 moles of C_8H_{18} react to give 18 moles of H_2O

So, 16.15 moles of C_8H_{18} react to give \frac{16.15}{2}\times 18=145.35 moles of H_2O

The moles of water produced are 145.35 moles.

<u>Solution for part (b) : Given,</u>

Mass of C_8H_{18} = 878 g

Molar mass of C_8H_{18} = 114 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of C_8H_{18}.

\text{ Moles of }C_8H_{18}=\frac{\text{ Mass of }C_8H_{18}}{\text{ Molar mass of }C_8H_{18}}=\frac{878g}{114g/mole}=7.702moles

Now we have to calculate the moles of O_2

The balanced chemical reaction is,

2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O

From the balanced reaction we conclude that

As, 2 moles of C_8H_{18} react with 25 moles of O_2

So, 7.702 moles of C_8H_{18} react with \frac{7.702}{2}\times 25=96.275 moles of O_2

Now we have to calculate the mass of O_2.

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

\text{ Mass of }O_2=(96.275moles)\times (32g/mole)=3080.8g

The mass of oxygen needed are 3080.8 grams.

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Answer:

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Explanation:

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As we know that

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