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3 years ago

Which of the following laboratory tools would be most appropriate for determining the volume of an irregular shape?

1 answer:

7 0

For an irregularly shaped objects, we use a GRADUATED CYLINDER to determine its volume by water displacement method. We fill the glassware first with water then measure the volume. Next, drop the object and read off the new volume. Calculate the difference between the volumes.

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which of the following methods is most appropriate to use to determine the number of different-colored components in a sample of

Andru [333]

The answer is paper chromatography using different solvents with a range of polarities as the mobile phase.

Paper chromatography- Low-molecular-mass molecules can be separated using paper chromatography based on how evenly they are distributed in the stationary and mobile phases. Paper chromatography is regarded as a potent analytical technique because of its low cost and the availability of numerous procedures for the separation of chemicals.

A small amount of a sample solution is poured onto a strip of chromatography paper in a paper chromatography experiment. After that, a solvent is used to suspend the chromatography paper. The sample solution's constituent components split out into bands of distinct hue as the solvent goes up the paper.

The speed of the chromatography process is influenced by the solvent's polarity. Therefore, we may conclude that all of the other components in the mixture move more quickly during the chromatography experiment if the solvent's polarity is increased.

Thus, answer is paper chromatography using different solvents with a range of polarities as the mobile phase.

To learn more about paper chromatography refer- brainly.com/question/1394204

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3 0

1 year ago

In the reaction Mg (s) + 2HCl (aq) -> H2 (g) + MgCl2 (aq), how many grams of hydrogen gas will be produced from 300 millilite

d1i1m1o1n [39]

Mg(s) + 2HCl(aq) = H₂(g) + MgCl₂(aq)

v=300 ml=0.3 l
c=4 mol/l

n(HCl)=vc

m(H₂)/M(H₂)=n(HCl)/2

m(H₂)=M(H₂)vc/2

m(H₂)=2.0g/mol·0.3 l · 4mol/l /2 = 1.2 g

1.2 grams of hydrogen gas will be produced

5 0

4 years ago

The thermal decomposition of N2O5 obeys first-order kinetics. At 45°C, a plot of ln[N2O5] versus t gives a slope of −6.40 × 10−4

finlep [7]

Answer:

<u>1,080 min</u>

<u></u>

Explanation:

A <em>first order reaction</em> follows the law:

$rate=k[A]$ , where [A] is the concentraion of the reactant A.

Equivalently:

$\dfrac{d[A]}{dt}=-k[A]$

Integrating:

$\dfrac{d[A]}{[A]}=-kdt$

$\ln \dfrac{[A]}{[A_o]}=-kt$

Half-life means [A]/[A₀] = 1/2, t = t½:

t½ = ln (2) / k

That means that the half-life is constant.

The slope of the plot of ln [N₂O₅] is -k. Then k is equal to 6.40 × 10⁻⁴ min⁻¹.

Thus, you can calculate t½:

t½ = ln(2) / 6.40 × 10⁻⁴ min⁻¹

t½ = 1,083 min.

Rounding to 3 significant figures, that is 1,080 min.

5 0

3 years ago

We’re did the pollution from acid rain and the factory eventually end up

Dimas [21]

Answer:

it ends up in our dirt and oceans/bodies of water

8 0

4 years ago

Read 2 more answers

Based on the three formulas shown, use one of them to solve for the purple yellow and red box and explain how you did it.

zysi [14]

P = 11.133 atm (purple)

T = -236.733 °C(yellow)

n = 0.174 mol(red)

<h3>Further explanation </h3>

Some of the laws regarding gas, can apply to ideal gas (volume expansion does not occur when the gas is heated),:

Boyle's law at constant T, P = 1 / V
Charles's law, at constant P, V = T
Avogadro's law, at constant P and T, V = n

So that the three laws can be combined into a single gas equation, the ideal gas equation

In general, the gas equation can be written

$\large {\boxed {\bold {PV = nRT}}}$

where

P = pressure, atm

V = volume, liter

n = number of moles

R = gas constant = 0.08206 L.atm / mol K

T = temperature, Kelvin

To choose the formula used, we refer to the data provided

Because the data provided are temperature, pressure, volume and moles, than we use the formula PV = nRT

Purple box

T= 10 +273.15 = 373.15 K

V=5.5 L

n=2 mol

$\tt P=\dfrac{nRT}{V}\\\\P=\dfrac{2\times 0.08205\times 373.15}{5.5}\\\\P=11.133~atm$

Yellow box

V=8.3 L

P=1.8 atm

n=5 mol

$\tt T=\dfrac{PV}{nR}\\\\T=\dfrac{1.8\times 8.3}{5\times 0.08205}\\\\T=36.42~K=-236.733^oC$

Red box

T = 12 + 273.15 = 285.15 K

V=3.4 L

P=1.2 atm

$\tt n=\dfrac{PV}{RT}\\\\n=\dfrac{1.2\times 3.4}{0.08205\times 285.15}\\\\n=0.174~mol$

3 0

3 years ago