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2 years ago

Post-Lab Questions

Chemistry

1 answer:

Svet_ta [14]2 years ago

7 0

Answer:

1. not enough dye was added to the drink.

The wrong dye was added to the drink

the water in the drink is evaporating

2. Changing the compound changes the absorbance behavior.

3. Measure the absorbance for the same solution in different cuvette sizes and find the y-intercept.

Explanation:

When the beverage company adds dye to the drink, there should be standard quantity added to the drink so that the color of the drink remains constant. When too much dye is added to the drink, the color will get dark brown or black. When the color of drink get lighter than green this means dye is not added in required quantity.

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How many grams of hydrogen are needed to react with 1.40 g of nitrogen to produce ammonia?

vazorg [7]

Answer:

Write a balanced chemical reaction:

N2 + 3H2 ==> 2NH3

Looking at the mole ratios in this balanced equation you can see it takes 3 moles H2 to make 2 moles NH3. So, next calculate the moles of NH3 represented by 1.80 g and then convert to moles of H2 needed:

moles of NH3 = 1.80 g x 1 mole/17 g = 0.106 moles NH3

Moles H2 needed = 0.106 moles NH3 x 3 moles H2/2 moles NH3 = 0.159 moles H2 needed

Grams H2 needed = 0.159 moles x 2 g/mole = 0.318 grams H2 needed

Explanation:

8 0

3 years ago

What pressure, in atmospheres, is exerted on the body of a diver if he is 38 ft below the surface of the water when atmospheric

Talja [164]

The pressure of diver = atmospheric pressure + water pressure

atmospheric pressure = 750 mmHg (as given) = 750 / 760 atm = 0.987 atm

Water pressure is

P = hρg

where

h = height of water = 38 ft

1 ft = 0.3048

38 ft = 11.58 m

ρ = density = 1000 Kg / m³

g = gravitational constant = 9.81 m/s2

P = 11.58 X 1000 X 9.81 = 113599.8 Kg / m s^2 Or N /m^2

1 N / m^2 = 1 pa = 9.869 X 10^-6 atm

P = 113599.8 Pa = 1.12 atm

Total pressure = 1.12 + 0.987 atm = 2.107 atm = 2.1 atm (two significant figures)

5 0

3 years ago

The half-life of radon-222 is 3.8 days. How many grams of radon-222 remain

Ne4ueva [31]

I think it’s B but not 100% sure

3 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

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3 years ago

This is the star at the center of the solar system around which the earth and other matter orbit

Fynjy0 [20]

The answer is the Sun.

Thank you

8 0

3 years ago