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wolverine [178]
2 years ago
8

Sometimes a new product is test-marketed. A limited amount of the product is released to a select group of users. Then the group

is observed carefully to see how the product performs. Which step might occur in the design process after a product is test-marketed and before large numbers are produced for sale?
Chemistry
1 answer:
Akimi4 [234]2 years ago
5 0

Answer:

Modification Step

Explanation:

The entire reason why test-marketing exists is to make sure that the product will be well received by the public. Once the test marketing concludes, the data from the test is analyzed. Then the company decides whether or not they will make changes on how the product looks, feels, performs, or even rebranding the product, all to make sure that the final consumer will like the product enough to want to purchase and continuously use the product. Which will in term increase profits for the company.

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A gas system contains 2.00 moles of O2 and CO2 gas, has an initial temperature of 25.0 oC and is under 1.00 atm of pressure. If
Ainat [17]

Answer:

V_2=52.2L

Explanation:

Hello there!

In this case, by bearing to to mind the given conditions, it is firstly possible to determine the initial volume of the closed system via the ideal gas equation:

PV=nRT\\\\V=\frac{2.00mol*0.08206\frac{atm*L}{mol*K}*298.15K}{1.00 atm} \\\\V=48.9L

Which is V1 in the Charles' law:

\frac{T_2}{V_2} =\frac{T_1}{V_1}

And of course, T1 is 298.15 (25+273.15). Therefore, by solving for V2 as the final volume, we obtain:

V_2=\frac{V_1T_2}{T_1}\\\\V_2=\frac{48.9L*(45+273.15)K}{(25+273.15)K}  \\\\V_2=52.2L

Best regards!

8 0
3 years ago
Oxygen gas is collected at a pressure of 123 kPa in a container which has a volume of 10.0L. What temperature must be maintained
Morgarella [4.7K]
Given:

P = 123 kPa
V = 10.0 L
n = 0.500 moles
T = ?

Assume that the gas ideally, thus, we can use the ideal gas equation:

PV = nRT

where R = 0.0821 L atm/mol K

123 kPa * 1 atm/101.325 kPa * 10.0 L = 0.500 moles * 0.0821 Latm/molK * T

solve for T 

T = 295.72 K<span />
5 0
3 years ago
Read 2 more answers
What is the correct noble gas configuration for strontium?
viktelen [127]
Kr 5s2 is the correct noble gas configuration for strontium
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3 years ago
What is a property in chemistry
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DescriptionA chemical property is any of a material's properties that becomes evident during, or after, a chemical reaction; that is, any quality that can be established only by changing a substance's chemical identity.
3 0
3 years ago
Tungsten (W) and chlorine (Cl) form a series of compounds with the following compositions:_______.
deff fn [24]

Answer:

WCl₂, WCl₄, WCl₅, WCl₆

Explanation:

Molar Mass of Tungsten = 184 g/mol

Mass of Chlorine = 35.5 g/mol

In the first compound;

Percentage of tungsten = 72.17 %

Upon solving;

72.17 % = 184

100 % = Total mass

Total mass of compound = 254.95g

Mass of chlorine = 254.95 - 184 = 70.95 (Dividing by 35.35; This is approximately 2 Chlorine atoms.

The Formular is WCl₂

In the second compound;

Percentage of tungsten = 56.45 %

Upon solving;

56.45 % = 184

100 % = Total mass

Total mass of compound = 325.95 g

Mass of chlorine = 325.95 - 184 = 141.95g (Dividing by 35.35; This is approximately 4 Chlorine atoms.

The Formular is WCl₄

In the third compound;

Percentage of tungsten = 50.91 %

Upon solving;

50.91 % = 184

100 % = Total mass

Total mass of compound = 361.42 g

Mass of chlorine =  361.42 - 184 = 177.42 (Dividing by 35.35; This is approximately 5 Chlorine atoms.

The Formular is WCl₅

In the fourth compound;

Percentage of tungsten = 46.39 %

Upon solving;

46.39 % = 184

100 % = Total mass

Total mass of compound = 396.64 g

Mass of chlorine = 396.64 - 184 = 212.64 (Dividing by 35.35; This is approximately 6 Chlorine atoms.

The Formular is WCl₆

4 0
3 years ago
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