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Rama09 [41]
2 years ago
8

Calculate the mass of a 0.297-mL sample of a liquid with a density of 0.930 g/mL.

Chemistry
1 answer:
tekilochka [14]2 years ago
8 0

Answer: 0.27621 g

Explanation:

0.297 ml *0.930 g/ml=0.27621 g

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it identifies reducing sugars (monosaccharide's and some disaccharides), which have free ketone or aldehyde functional group

Explanation:

It turns from turquoise to yellow or orange when it reacts with reducing sugars.

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2 years ago
Which of the following best states the difference between an alcohol and an ether?
goldenfox [79]
I'd say it would be
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3 years ago
Name the organic compound of ch3chohch2ch3
snow_lady [41]

Answer:

Flavouring agent 2-Butanol, or sec-butanol, is an organic compound with formula CH3CH(OH)CH2CH3.

4 0
3 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

2)

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

3)

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
To plot the calibration curve, you need to prepare iron solutions with known concentrations and measure their absorbance.
kap26 [50]

To plot the calibration curve, you need to prepare iron solutions with known concentrations and measure their absorbance. You need to pipet 0 mL of the diluted solution to have 0.00 mg of iron.

In spectrophotometry, to plot the calibration curve, you need to prepare solutions with known concentrations and measure their absorbance.

We have a standard iron solution with a concentration of 0.2500g/L of pure iron (C₁). We pipet 25.00mL (V₁) of this standard iron solution into a 500mL (V₂) volumetric flask and dilute up to the mark with distilled water.

We can calculate the concentration of the diluted solution (C₂) using the dilution rule.

C_1 \times V_1 = C_2 \times V_2\\\\C_2 = \frac{C_1 \times V_1}{V_2}  = \frac{0.2500 g/L \times 25.00 mL}{500 mL} = 0.0125 g/L

Then, if we wanted to prepare the blank, that is, the solution that contains the same matrix but not the analyte, and whose concentration in iron is 0.00 mg/L, we wouldn't pipet any of the diluted solution.

To plot the calibration curve, you need to prepare iron solutions with known concentrations and measure their absorbance. You need to pipet 0 mL of the diluted solution to have 0.00 mg of iron.

Learn  more: brainly.com/question/24195565

8 0
2 years ago
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