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Lostsunrise [7]
3 years ago
15

. Chemical manufacturers must present which Information on the product's label?

Engineering
1 answer:
STatiana [176]3 years ago
5 0

Answer: A) Product identifier

Explanation:

A product identifier is a means, name or number used to identify hazardous and dangerous chemicals on a label or in the SDS( safety data sheet). It provides a special means which helps the user to identify the chemical. Any product identifier used should permit a cross-references to be made among the listed hazardous chemicals required in the written hazard communication program, and the label including SDS(safety data sheet).

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The crash rate per mile is.
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Answer:

75 percent

Explanation:

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3 years ago
An Otto cycle engine is analyzed using the air standard method. Given the conditions at state 1, compression ratio (r), and pres
My name is Ann [436]

Answer:

A)  222.58 kJ / kg

B)  0.8897 M^3/ kg

c)  0.7737 m^3/kg

D)  746.542 k

E)  536.017 kj/kg

efficiency = 58% ( approximately )

Explanation:

Given Data :

Gas constant (R) =  0.287 kJ/ kg.K

T1 = 310 k

P1 ( Kpa ) = 100

r = 11.5 ( compression ratio )

rp = 1.95 ( pressure ratio )

A ) specific internal energy at state 1

 = Cv*T1 =  0.718 * 310 = 222.58 kJ / kg

B) Relative specific volume at state 1

= P1*V1 = R*T1 ( ideal gas equation )

V1 = R*T1 / P1 = (0.287* 10^3*310 ) / 100 * 10^3

V1 = 88.97 / 100 = 0.8897 M^3/ kg

C ) relative specific volume at state 2

Applying  r ( compression ratio) = V1 / V2

11.5 = 0.8897 / V2

V2 = 0.8897 / 11.5 = 0.7737 m^3/kg

D) The temperature (k) at state 2

since the process is an Isentropic process we will apply the p-v-t relation

\frac{T1}{T2} = (\frac{V1}{V2}^{n-1}  ) = (\frac{P2}{P1} )^{\frac{n-1}{n} }

hence T2 = 9^{1.4-1} * 310 = 2.4082 * 310 = 746.542 k

e) specific internal energy at state 2

= Cv*T2 = 0.718  * 746.542 = 536.017 kj/kg

efficiency = output /input = 390.3511 / 667.5448 ≈ 58%

attached is a free hand diagram of an Otto cycle is attached below

3 0
3 years ago
Basic concepts surrounding electrical circuitry?​
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Hopefully that helps you out and is this for history or science?

3 0
2 years ago
Steam at 75 kPa and 8 percent quality is contained in a spring-loaded piston–cylinder device, as shown in Figure, with an initia
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The heat transferred to and the work produced by the steam during this process  is 13781.618 kJ/kg

<h3>​How to calcultae the heat?</h3>

The Net Change in Enthalpy will be:

= m ( h2 - h1 ) = 11.216 ( 1755.405 - 566.78 ) = 13331.618 kJ/kg

Work Done (Area Under PV curve) = 1/2 x (P1 + P2) x ( V1 - V2)

= 1/2 x ( 75 + 225) x (5 - 2)

W = 450 KJ

From the First Law of Thermodynamics, Q = U + W

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= 13331.618 + 450

Q = 13781.618 kJ/kg

Learn more about heat on:

brainly.com/question/13439286

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