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2 years ago

Consider the two facts below:

Chemistry

1 answer:

OLEGan [10]2 years ago

4 0

Answer:

A. There is more dissolved oxygen in colder waters than in warm water.

D. If ocean temperature rise, then the risk to the fish population increases.

Explanation:

Conclusion that can be drawn from the two facts stated above:

*Dissolved oxygen is essential nutrient for fish survival in their aquatic habitat.

*Dissolved oxygen would decrease as the temperature of aquatic habit rises, and vice versa.

*Fishes, therefore, would thrive best in colder waters than warmer waters.

The following are scenarios that can be explained by the facts given and conclusions arrived:

A. There is more dissolved oxygen in colder waters than in warm water (solubility of gases decreases with increase in temperature)

D. If ocean temperature rise, then the risk to the fish population increases (fishes will thrive best in colder waters where dissolved oxygen is readily available).

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Describe how you would prepare approximately 2 l of 0.050 0 m boric acid, b(oh)3.

Elenna [48]

The given concentration of boric acid = 0.0500 M

Required volume of the solution = 2 L

Molarity is the moles of solute present per liter solution. So 0.0500 M boric acid has 0.0500 mol boric acid present in 1 L solution.

Calculating the moles of 0.0500 M boric acid present in 2 L solution:

$2 L * \frac{0.0500 mol B(OH)_{3} }{1 L} = 0.100 mol B(OH)_{3}$

Converting moles of boric acid to mass:

$0.100 mol B(OH)_{3} * \frac{61.83 g}{mol B(OH)_{3}} = 6.183 g$

Therefore, 6.183 g boric acid when dissolved and made up to 2 L with distilled water gives 0.0500 M solution.

5 0

2 years ago

Suppose that a metal oxide of formula m2o3 were soluble in water. what would be the major product or products of dissolving the

V125BC [204]

Meta oxides are compounds that are formed by reaction of metals with oxygen. If these compounds are placed in water, the ionic components of this substance will dissociate.

The dissociation of metal oxides in water will likely form,

2M³⁺ + 3O²⁻

6 0

2 years ago

What is the percent yield of this reaction if 22.o g of Mgl2 is produced by the reaction of 25.0 g of Mg with 25.0 g of l2?

expeople1 [14]

% yield = 80.719

<h3>Further explanation</h3>

Given

22.0 g of Mgl₂

25.0 g of Mg

25.0 g of l₂

Required

The percent yield

Solution

Reaction

Mg + I₂⇒ MgI₂

mol Mg = 25 g : 24.305 g/mol = 1.029

mol I₂ = 25 g : 253.809 g/mol = 0.098

Limiting reactant = I₂

Excess reactant = Mg

mol MgI₂ based on I₂, so mol MgI₂ = 0.098

Mass MgI₂ (theoretical):

= mol x MW

= 0.098 x 278.114

= 27.255 g

% yield = (actual/theoretical) x 100%

% yield = (22 / 27.255) x 100%

% yield = 80.719

8 0

1 year ago

An amount of solid barium chloride, 20.8 g, is dissolved in 100 g water in a coffee-cup calorimeter by the reaction: BaCl2 (s) 

mamaluj [8]

Answer : The enthalpy change during the reaction is -6.48 kJ/mole

Explanation :

First we have to calculate the heat gained by the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

m = mass of water = 100 g

c = specific heat = $4.04J/g^oC$

$T_{final}$ = final temperature = $26.6^oC$

$T_{initial}$ = initial temperature = $25.0^oC$

Now put all the given values in the above formula, we get:

$q=100g\times 4.04J/g^oC\times (26.6-25.0)^oC$

$q=646.4J$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles barium chloride = $\frac{\text{Mass of barium chloride}}{\text{Molar mass of barium chloride}}=\frac{20.8g}{208.23g/mol}=0.0998mole$