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3 years ago

Why is it better to use the metric system, rather than the English system, in scientific measurement?

Chemistry

2 answers:

nasty-shy [4]3 years ago

6 0

Answer:

Explanation:

Hello,

English system is indeed based on the metric system because they look for reference stuff to get the measurements, for instance, a foot equals 30.48 cm, but a common 30-cm rule measures every 1 cm, so the metric system also provides more accurate measurements. An other example lies on the measurement of mass, a pound which equals 453.59 g is appropriated to measure middle amounts of masses, but for example, in qualitative analysis we tend to measure amounts below about 0.10 g (0.022 lb) so it is easier for us in science to use the metric system in addition to the straightforwardness of the reporting data in 10-based notation (scientific notation).

Best regards.

frez [133]3 years ago

3 0

The metric system is based on powers of 10 so it is much easier to convert units, often just by moving the decimal point.

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For green plant cells, which are the substances that result from photosynthesis that are used in cellular respiration? A. oxygen

strojnjashka [21]

Answer:

B. oxygen and glucose

Explanation:

Oxygen for respiration and glucose for energy utilization

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3 years ago

Consider the reaction pathway graph below.

san4es73 [151]

Hello!

If the reaction could be made to go faster B. The hill would be shorter.

The "hill" is called Activation Energy. It is the energy of the activated intermediate compound needed to complete the reaction. Catalysts are substances that don't intervene in the reaction which speed up the rate of a given reaction by lowering the Activation Energy giving alternate reaction pathways with lower-energy intermediates. If the reaction could be made to go faster with the use of a catalyst, then the Activation Energy would be lower and the "hill" would be shorter.

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3 years ago

One hundred fifty joules of heat are removed from a heat reservoir at a temperature of 150 K. What is the entropy change of the

Sidana [21]

Answer:

ΔS surrounding (entropy change of the reservoir) = -1 J/K

Explanation:

Given:

Change in heat (ΔH) = 150 joules

Temperature (T) = 150 K

Find:

ΔS surrounding (entropy change of the reservoir)

Computation:

ΔS surrounding (entropy change of the reservoir) = - ΔH / T

ΔS surrounding (entropy change of the reservoir) = - 150 / 150

ΔS surrounding (entropy change of the reservoir) = -1 J/K

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3 years ago

A 2.350×10−2 M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving a sample of NaCl in water and then b

sveta [45]

Answer:

Part A: m = 0.02356 mol/kg = 0.02356 m

Part B: Xsolute = 4.243×10⁻⁴

Part C: % m/m = 0.1376%

Part D: ppm = 1,376 ppm

Explanation:

1. Data:

a) M = 2.350×10⁻² M

b) V sol = 1.000 L

c) V H₂O = 994.4 mL = 0.9944 L

d) d H₂O = 0.9982 g/mL

2. Formulae

M = n solute / V sol (L)
m = n solute / Kg solvent
X solute = n solute / N total
% m/m = (mass of solute / mass of solution) × 100
ppm = (mass of solute / mass of solution) × 1,000,000
density = mass in grams / volume in mL

3. Solution

Part A: Calculate the molality of the salt solution.

m = n solute / Kg solvent

i) M = n solute / V sol (L) ⇒ n solute = M × V sol (L)

⇒ n solute = M = 2.350×10⁻² M × 1.000 L = M = 2.350×10⁻² mol

ii) density H₂O = mass H₂O / volume H₂O

⇒ mass H₂O = density H₂O × volume H₂O

⇒ mass H₂O = 0.9982 g/mL × 999.4 mL = 997.6 g

iii) kg H₂O = 997.6 g / (1,000 g/Kg) = 0.9976 kg

iv) m = 2.350×10⁻² mol / 0.9976 kg = 0.02356 mol/kg = 0.02356 m

Part B: Calculate the mole fraction of salt in this solution.

X solute = n solute / N total

i) n solute = 2.350×10⁻² mol

ii) n solvent = n H₂O = mass H₂O in grams/ molar mass H₂O

⇒ 997.6 g / 18.015 g/mol = 55.38 mol

iii) X solute = 2.350×10⁻² mol / 55.38 mol = 4.243×10⁻⁴

Part C: Calculate the concentration of the salt solution in percent by mass.

% m/m = (mass of solute / mass of solution) × 100

i) molar mass = mass in grams / molar mass

⇒ mass of solute = mass of NaCl = n solute × molar mass NaCl

⇒ mass of solute = 2.350×10⁻² mol × 58.44 g/mol = 1.373 g

ii) % m/m = (1.373 g / 997.6 g) × 100 = 0.1376%

Part D: Calculate the concentration of the salt solution in parts per million.

ppm = (mass of solute / mass of solution) × 1,000,000

i) ppm = ( (1.373 g / 997.6 g) × 1,000,000 = 1,376 ppm

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Alex73 [517]

An atom is the smallest unit if a matter that has properties of an element.
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