_______ are hollow spaces in the earths crust?.

Lastly, Snape thinks we should try one more calculation.

EleoNora [17]

Answer:

E) 0.90

Explanation:

In TLC (Thin-Layer-Chromatography) retention factor RF is defined as the ratio between distance of the sample and the solvent front. RF is very important in chemistry to know the composition of any sample by comparison.

In the problem, as the sample has a distance of 0,20cm from the solvent front, the distance of the sample is:

2,0cm - 0,20 cm =<em> 1,8 cm</em>. Thus, RF is:

RF = 1,8cm / 2,0cm = 0,90

<em>E) 0,90</em>

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I hope it helps!

3 0

2 years ago

What is the molecular formula for a compound that is 44.87% potassium, 36.7%

Phoenix [80]

Answer:

Molecular formula: S4K8O16 empirical formula: SK2O4

Explanation:

First we find the moles of each by first finding grams (using the percent) and then using stoichiometry to convert into moles:

Sulfur: 696 *.18 = 125.28grams S* $\frac{1 mole S}{32.065 g S} = 3.907 moles S$

Potassium: 696 *.4487 = 312.2952 * $\frac{1 mole K}{39.08 g K}$ = 7.99117 mole K

Oxygen: 696 * .367 = 255.432 * $\frac{1 mol O}{16g O}$ = 15.9654 mole O

Then we divide each value by the atom with the smallest number of moles to find the mole ratio:

3.907/3.907= 1

7.99117 mole K/ 3.907= 2.043

15.9654 mole O/ 3.907= 4.08

The empirical formula is SK2O4

To find the molecular formula, we divide the mass given (696) by the mass of the empirical formula (174.22) to get 4. We then divide each atom by 4.

Molecular formula: S4K8O16

5 0

2 years ago

our throat is dry, and you want the last cough drop in the box to last a long time in your mouth. What should you do? Keep the c

Ann [662]

You should keep the cough drop whole. This maintains the largest surface-to-volume ratio and slows the dissolution of the cough drop.

<h3>Delay of drug action</h3>

The action of the drug can be delayed to either reduce adverse effect if the drugs to the body or to enhance its therapeutic purposes.

Because the throat is dry, the therapeutic purposes of the cough drop would only be achieved if it's dissolution is delayed in the mouth.

To do so, the cough drop should be kept whole to maintain the largest surface-to-volume ratio. This would slow the dissolution of the cough drop.

Learn more about drug here:

brainly.com/question/26254731

8 0

1 year ago

P O R F A V O R

ale4655 [162]

Answer:

a. 45×10³ kg

b. 1.25 kg

c. 5443200 s

d. 2.69×10⁻⁴ m/s²

e. 2.57×10⁻⁶N

f. 7.48×10⁻³ m /s²

g. 2.45 Pa

h. 10 m/s

Explanation:

The SI units are: kg, m, s, N, K, A, Pa, J and cd

a. 1 g is the mass for 1 cm³. We convert the m³ to cm³

45 m³. 1×10⁶ cm³ / 1 m³ = 45×10⁶ cm³

By the way, 45×10⁶ cm³ = 45×10⁶ g

We convert the g to kg → 45×10⁶ g . 1 kg / 1000 g = 45×10³ kg

b. As 1 g = 1 cm³, we convert the cm³ to g and then, the g to kg

1250 cm³ = 1250 g → 1250 g . 1kg / 1000 g = 1.25 kg

c. 1 day has 24 hours; 1 hour has 60 minutes; 1 minute has 60 seconds

1 hour has 3600 s. Then 24 h . 3600 s / 1 h = 86400 s

86400 s/d. 63 d = 5443200 s

d. 1 min² = 3600 s²

97 cm / 3600 s² = 0.0269 cm/s² / 100 = 2.69×10⁻⁴ m/s²

e. 927 g.cm / min² / 3600 s² = 0.2575 g.cm/s² → dyn

We need to convert dyn to N

1 dyn = 10⁻⁵N → 0.2575 dyn . 10⁻⁵N / 1dyn = 2.57×10⁻⁶N

f. 1 m/s² = 12960 km/h²

12960 km/h² . 1 m/s² / 97 km/h² = 7.48×10⁻³ m /s²

g. 2500 g/cm² . 1kg / 1000 g = 2.5 kg/cm²

1 Pa = 1.02kg/cm²

2.5 kg/cm² . 1 Pa / 1.02 kg/cm² = 2.45 Pa

h. 1 h = 3600 s

36 km / 3600 s = 0.01 km /s → 0.01 km . 1000 m / 1 km = 10

= 10 m/s

5 0

2 years ago

Calculate the fraction of atoms in a sample of argon gas at 400 K that have an energy of 10.0 kJ or greater.

nikdorinn [45]

Answer:

The answer to this can be arrived at by clculating the mole fraction of atoms higher than the activation energy of 10.0 kJ by pluging in the values given into the Arrhenius equation. The answer to this is 20.22 moles of Argon have energy equal to or greater than 10.0 kJ

Explanation:

From Arrhenius equation showing the temperature dependence of reaction rates.

$K = Ae^{\frac{Ea}{RT} }$ where

k = rate constant

A = Frequency or pre-exponential factor

Ea = energy of activation

R = The universal gas constant

T = Kelvin absolute temperature

we have

$f = e^{\frac{Ea}{RT} }$

Where

f = fraction of collision with energy higher than the activation energy

Ea = activation energy = 10.0kJ = 10000J

R = universal gas constant = 8.31 J/mol.K

T = Absolute temperature in Kelvin = 400K

In the Arrhenius equation k = Ae^(-Ea/RT), the factor A is the frequency factor and the component e^(-Ea/RT) is the portion of possible collisions with high enough energy for a reaction to occur at the a specified temperature

Plugging in the values into the equation relating f to activation energy we get

$f = e^{\frac{10000J}{(8.31J/((mol)(K)))(400K)} }$ or f = $e^{3.01}$ = 20.22 moles of argon have an energy of 10.0 kJ or greater

5 0

2 years ago