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2 years ago

Will mark BRAINLIEST

Chemistry

1 answer:

likoan [24]2 years ago

4 0

Answer:

A and D

Explanation:

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Why would scientists change their ideas? *<br> Need help

n200080 [17]

Answer:

If another scientist give better reasons;if they tested their hypothesis and it was wrong; if someone proved whatever the idea was to be wrong

Explanation:

I mean the list can go on at this point

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2 years ago

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2 years ago

A fluid occupying has a mass of 4mg. Calculate its density and specific volume in SI, EE, and BG units.

kondaur [170]

The question is incomplete, complete question is:

A fluid occupying $3.2 m^3$ of volume has a mass of 4 Mg. Calculate its density and specific volume in SI, EE, and BG units.

Explanation:

1) Mass of liquid = m = 4 Mg = 4 × 1,000 kg = 4,000 kg

(1 Mg = 1000 kg)

Volume of the fluid = V = $3.2 m^3$

Density of the fluid = D

$D=\frac{m}{V}=\frac{4,000 kg}{3.2 m^3}=1,250 kg/m^3$

Specific volume is the reciprocal of the density :

$V_{specific}=\frac{1}{Density}$

Specific volume of the fluid = $S_v$

$S_v=\frac{1}{D}=\frac{1}{1,250 kg/m^3}=0.0008 m^3/kg$

Density of the fluid in English Engineering units = D $(lb/ft^3)$

1 kg = 2.20462 lb

1 m = 3.280 ft

$D=\frac[1,250\times 2.20462 lb}{(3.280 ft)^3=78.95 lb/ft^3$

Specific volume of the fluid :

$=\frac{1}{78.95 lb/ft^3}=0.0127 ft^3/lb$

Density of the fluid in British Gravitational System units = D $(slug/ft^3)$

1 kg = 0.06852 slug

1 m = 3.280 ft

$D=\frac[1,250\times 0.0685218 slug}{(3.280 ft)^3=2.43 slug/ft^3$

Specific volume of the fluid :

$=\frac{1}{2.43 slug/ft^3}=0.412 ft^3/slug$

7 0

3 years ago

(20 points) i) An absorption intensity of 1.00 (in arbitrary units) is observed for the maximum peak of the 1:2:1 triplet of the

laiz [17]

Answer:

Concentration of ethanol required = 48.476 M

Explanation:

Given that:

the absorption intensity = 1.00

Molarity of ethanol = 1M

NMR instrument used = 160 MHz

Temperature used = 300 K

The required concentration of ethanol can be determined as follows:

$= ( 1 \ M \times \dfrac{160\ MHz }{450 \ MHz}) \times \dfrac{300 \ K}{2.2\ K}$

$= ( 1 \ M \times 0.3555 ) \times136.36}$

= 48.476 M

5 0

3 years ago