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mariarad [96]
3 years ago
13

• 1. An object has a mass of 570g and a volume of 2280ml. Calculate its density. 250 kg/m3

Chemistry
1 answer:
Dima020 [189]3 years ago
7 0

Answer:

Density=mass/volume

= 570/2280

=0.25g/ml

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1.What is adaptation and why is it important?
IrinaVladis [17]

Answer:

All organisms need to adapt to their habitat to be able to survive. This means adapting to be able to survive the climatic conditions of the ecosystem, predators, and other species that compete for the same food and space.

Explanation:

4 0
3 years ago
The Ka1 value for oxalic acid is 5.9 x10-2 , and the Ka2 value is 4.6 x 10-5 . What are the values of Kb1 and Kb2 of the oxalate
andre [41]

Answer:

2.17x10⁻¹⁰ = Kb1

1.69x10⁻¹³ = Kb2

Explanation:

Oxalic acid, C₂O₄H₂, has two intercambiable protons, its equilibriums are:

C₂O₄H₂ ⇄ C₂O₄H⁻ + H⁺ Ka1 = 5.9x10⁻²

C₂O₄H⁻ ⇄ C₂O₄²⁻ + H⁺ Ka2 = 4.6x10⁻⁵

Oxalate ion, C₂O₄²⁻, has as equilibriums:

C₂O₄²⁻ + H₂O ⇄ C₂O₄H⁻ + OH⁻ Kb1

C₂O₄H⁻ + H₂O ⇄ C₂O₄H₂ + OH⁻ Kb2

Also, you can know: KaₓKb = Kw

<em>Where Kw is 1x10⁻¹⁴</em>

Thus:

Kw = Kb2ₓKa1

1x10⁻¹⁴ =Kb2ₓ4.6x10⁻⁵

<h3>2.17x10⁻¹⁰ = Kb1</h3>

And:

Kw = Kb1ₓKa2

1x10⁻¹⁴ =Kb1ₓ5.9x10⁻²

<h3>1.69x10⁻¹³ = Kb1</h3>

<em>That is because the inverse reaction of, for example, Ka1:</em>

<em>C₂O₄H⁻ + H⁺ ⇄ C₂O₄H₂ K = 1 / Ka1</em>

<em>+ H₂O ⇄ H⁺ + OH⁻ K = Kw = 1x10⁻¹⁴</em>

<em>= </em>

<em>C₂O₄H⁻ + H₂O ⇄ C₂O₄H₂ + OH⁻ Kb2 = Kw × 1/Ka1</em>

6 0
3 years ago
20 Points for urgency. <br> Determine the [OH-] for the following.<br> (Shown in image)
vredina [299]
1.pOH+pH=14
pOH+1*10^=14
pOH=13.9999
3 0
3 years ago
A 10.0 g ice cube is placed into 250 g of water with an initial temperature of 20.0 C. If the water drops to a temperature of 16
Radda [10]

the mass of ice taken = 10 g

the mass of water = 250 g

initial temperature of water = 20 C

the final temperature of water = 16. 8 C

specific heat of water = 4.18 J/g*K

the heat absorbed by ice to melt = heat loss by water

heat loss by water = mass X specific heat of water X change in temperature

heat loss by water = 250 X 4.18 X (20-16.8) = 3344 Joules

heat gained by ice = 3344 J

heat gained by ice = enthalpy of fusion X moles of ice

moles of ice = mass / molar mass = 10 / 18 = 0.56 moles

enthalpy of fusion = 3344 / 0.56 = 5971.43 J / mole

3 0
3 years ago
A balloon containing 0.0400 mol of a gas with a volume of 500 mL was expanded to 1.00 L. Answer the questions and round answers
grandymaker [24]

the number of moles 0.08

8 0
3 years ago
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