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4 years ago

What are the requirements for chemical labels

1 answer:

3 0

A chemical label should include: a product identifier, signal word, hazard statement, precautionary statement,pictorgrams and supplier information

Explanation:

The product identifier : positioned at the upper left end corner of the label gives the identity of the hazardous chemical by appropriate terms. i.e. the name and code

Signal word: The "Danger" and "WARNING" words. The former is used for a more severe hazard where as the former is for a less severe hazard

Hazard statement : gives a description of the nature and degree of hazard of the chemical

Precautionary statements : Put for workers and users on ways to minimize exposure and lower risks in event of contact with the harmful chemical

Supplier information : It has the chemical manufacturer details such as name, location and telephone number.

Pictograms : Normally symbols with red borders for visual illustration of the hazard chemical.

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Chemical labels : brainly.com/question/3115518

Keywords : chemical, labels

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in 1984 don, disk that stayed aloft for 16.7 seonds supsuppose cain ran up a staricase during this time, reachig a height of 18.

laila [671]

Answer

given,

time of disk stayed a loft = 16.7 sec

height reached = 18.4 m

mass = 72 kg

now work done is equal to change in potential energy

$W = - (U_f - U_i)$

$W = - (mgh - 0)$

$W = - (72\times (-9.8) \times 18.4)$

W = 12983.04 J

Positive sign means that work is in the direction of motion.

$Power = \dfrac{Work\ done}{time}$

$Power = \dfrac{W}{t}$

$Power = \dfrac{12983.04}{16.7}$

P = 777.42 Watts

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How does an acid–base indicator react when placed in an acidic solution?

s344n2d4d5 [400]

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4 years ago

On a transverse wave, the crest isa. the highest part of the wave.

Greeley [361]

I think it is C) the resting position of the wave.

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3 years ago

A crate with a mass of 175.5 kg is suspended from the end of a uniform boom with a mass of 94.7 kg. The upper end of the boom is

kotykmax [81]

323.5 N is the tension in the cable.

Given

Mass of crate(M) = 175.5 kg

Mass of boom(m) = 94.7 kg

The tension, T in the cable can be calculated by taking moments of force about the central point of marked X.

The Angle of the boom with the horizontal can be calculated by

tanθ = 5/10

θ = tan⁻¹(5/10) = 26.56°

Angle of the boom with horizontal is 26.56°

The angle of cable with horizontal can be calculated by

tan B = 4/10

B = tan⁻¹(4/10) = 21.80°

Angle of cable with horizontal is 21.80°

Taking moments of force about the point X

(Mcosθ + mcosθ) 0.5 = T(sin(θ +B)1

(175.5 × cos 26.56 + 94.7 × cos 26.56 )× 0.5 = T (sin(26.56 + 21.80) X 1

By calculating, we get

Tension(T) = 241.68/0.747

Tension(T) = 323.5 N

Hence, 323.5 N is the tension in the cable.

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