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2 years ago

14

Define the term standardization

1 answer:

trapecia [35]2 years ago

6 0

Explanation:

Standardization is the process of creating protocols to guide the creation of a good or service based on the consensus of all the relevant parties in the industry. ... Standardization also helps in ensuring the safety, interoperability, and compatibility of goods produced.

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When the volume of a gas is changed from 3.6 L to 15.5 L, the temperature will change from ______ C to 87 C.

harkovskaia [24]

V1/T1=V2/T2
(15.5)/(360K)=(3.6)/(T2)
T2=83.61290K
T2=-189.3871 degrees Celsius

5 0

3 years ago

Read 2 more answers

Calculate the value of the diffusion coefficient D (in m2/s) at 547°C for the diffusion of some species in a metal; assume that

svetlana [45]

Answer : The value of diffusion coefficient is, $2.97\times 10^{-16}m^2/s$

Explanation :

Formula used :

$D=D_o\times \exp \left(-\frac{Q_d}{RT}\right )$

where,

D = diffusion coefficient = ?

$D_o$ = $5.6\times 10^{-5}m^2/s$

$Q_d$ = $177kJ/mol$

R = gas constant = 8.314 J/mol.K

T = temperature = $547^oC=273+547=820K$

Now put all the given values in the above formula, we get:

$D=5.6\times 10^{-5}m^2/s\times \exp \left(-\frac{177kJ/mol}{(8.314J/mol.K)\times (820K)}\right )$

$D=2.97\times 10^{-16}m^2/s$

Thus, the value of diffusion coefficient is, $2.97\times 10^{-16}m^2/s$

6 0

3 years ago

Renewable energy resources include *

vitfil [10]

It’s A, so welcome my friend

3 0

2 years ago

Calculate the number of moles of each compound, given the number of molecules.

anastassius [24]

Answer:

1. $2.46\times 10^{21}$ molecules of CO₂

2. 10⁴ molecules of H₂O

3. 8.75×10³² molecules of C₆H₁₂O₆

Explanation:

1. $2.46\times 10^{21}$ molecules of CO₂

2. 10⁴ molecules of H₂O

3. 8.75×10³² molecules of C₆H₁₂O₆

5 0

3 years ago

What quantity in moles of LiBr are in 66.1 grams of LiBr?

anygoal [31]

Molar mass of LiBr

7+80=87g/mol

$\\ \sf\hookrightarrow No\:of\;moles=\dfrac{Given\:mass}{Molar\:mass}$

$\\ \sf\hookrightarrow No\:of\:moles=\dfrac{66.1}{87}$

$\\ \sf\hookrightarrow No\:of\:moles=0.75mol$

5 0

2 years ago