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Rasek [7]
3 years ago
11

Leopard seals are animals adapted to survive in the freezing conditions of Antarctica. Individual leopard seals vary in their di

fferent traits.
Which of the following variations would give a leopard seal the greatest chance of surviving in its harsh environment?
A. More spots on its coat. B. Thicker layer of blubber
C. More elongated head
D. Wider, flatter teeth
Chemistry
1 answer:
andreev551 [17]3 years ago
6 0

Answer:

B. Thicker layer of blubber

Explanation:

For the leopard seals to survives this harsh environment, it must have a thicker layer of blubber

Blubber is a thick layer of fat. It is called the adipose tissues and found in most marine organisms.

  • This layer helps in insulating the body against heat loss.
  • By so doing, the animal is able to conserve internal heat.
  • They have low thermal conductivity and do not easily lose heat or gain heat as such.
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What are the reactants and products in a chemical reaction?
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The reactant is the chemical reaction of 2 or more different elements. 
The product is what is formed in a chemical reaction. 

Example: H + O ---->  H2O 
H + O is the reactants, and H2O is the products. 
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When Al(NO3)3 is separated into its ions, how is it written?
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Al (+3)
NO3 (-1)
So basically you need 3 NO3 and one Al to produce Al (NO3)3
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A solution is made by adding 50.0 ml of 0.200 m acetic acid (ka = 1.8 x 10–5) to 50.0 ml of 1.00 x 10–3m hcl. (a) calculate the
Irina18 [472]

Answer:

Final pH of the solution: 2.79.

Explanation:

What's in the solution after mixing?

\displaystyle c = \frac{n}{V},

where

  • c is the concentration of the solute,
  • n is the number of moles of the solute, and
  • V is the volume of the solution.

V(\text{Final}) = 0.050 \;\textbf{L} + 0.050\;\textbf{L} = 0.100\;\textbf{L}.

Acetic (ethanoic) acid:

\displaystyle \begin{aligned}n &= c(\text{Before})\cdot V(\text{Before}) \\&= 0.050\;\text{L} \times 0.200\;\text{mol}\cdot\text{L}^{-1}\\ &= 0.0100\;\text{mol}\end{aligned}.

\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{0.0100\;\text{mol}}{0.100\;\text{L}}\\ &= 0.100\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 0.100\;\text{M}\end{aligned}.

Hydrochloric acid HCl:

\begin{aligned}n &= c(\text{Before})\cdot V(\text{Before})\\ &= 0.050\;\text{L} \times 1.00\times 10^{-3}\;\text{mol}\cdot\text{L}^{-1}\\ &= 5.00\times 10^{-5}\;\text{mol}\end{aligned}.

\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{5.00\times 10^{-5}\;\text{mol}}{0.100\;\text{L}}\\ &= 5.00\times 10^{-4}\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 5.00\times 10^{-4}\;\text{M}\end{aligned}.

HCl is a strong acid. It will completely dissociate in water to produce H⁺. The H⁺ concentration in the solution before acetic acid dissociates shall also be 5.00\times 10^{-4}\;\text{M}.

The Ka value of acetic acid is considerably small. Acetic acid is a weak acid and will dissociate only partially when dissolved. Construct a RICE table to predict the portion of acetic acid that will dissociate. Let the change in acetic acid concentration be -x\;\text{M}. x > 0.

\begin{array}{c|ccccc}\textbf{R}&\text{CH}_3\text{COOH}\;(aq) &\rightleftharpoons &\text{CH}_3\text{COO}^{-}\;(aq) &+& \text{H}^{+}\;(aq)\\\textbf{I}&0.100\;\text{M} & & & & 5.00\times 10^{-4}\;\text{M}\\\textbf{C}&-x\;\text{M} & & +x\;\text{M} & & +x\;\text{M} \\ \textbf{E}&0.100\;\text{M}-x\;\text{M} & & x\;\text{M} & & 5.00\times 10^{-4}\;\text{M} + x\;\text{M}\end{array}.

\displaystyle K_a = \frac{[\text{CH}_3\text{COO}^{-}\;(aq)]\cdot[\text{H}^{+}\;(aq)]}{[\text{CH}_3\text{COOH}\;(aq)]} = \frac{x\cdot(x + 5.00\times 10^{-4})}{0.100 - x}.

Rewrite as a quadratic equation and solve for x:

x\cdot(x + 5.00\times 10^{-4}) = (1.8\times 10^{-5} )\cdot (0.100 - x)

x\approx 0.00111.

The pH of a solution depends on its H⁺ concentration.

At equilibrium

[\text{H}^{+}\;(aq)] = 5.00\times 10^{-4}\;\text{M} + x\;\text{M} = 0.00161\;\text{M}.

\text{pH} = -\log{[\text{H}^{+}]} = 2.79.

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3 years ago
The fuel used in many disposable lighters is liquid butane, C4H10. Butane has a molecular weight of 58.1 grams in one mole. How
BARSIC [14]
<h3>Answer:</h3>

              6.21 × 10²² Carbon Atoms

<h3>Solution:</h3>

Data Given:

                 Mass of Butane (C₄H₁₀)  =  1.50 g

                 M.Mass of Butane  =  58.1 g.mol⁻¹

Step 1: Calculate Moles of Butane as,

                 Moles  =  Mass ÷ M.Mass

Putting values,

                 Moles  =  1.50 g ÷ 58.1 g.mol⁻¹

                 Moles  =  0.0258 mol

Step 2: Calculate number of Butane Molecules;

As 1 mole of any substance contains 6.022 × 10²³ particles (Avogadro's Number) then the relation for Moles and Number of Butane Molecules can be written as,

            Moles  =  Number of C₄H₁₀ Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Solving for Number of Butane molecules,

             Number of C₄H₁₀ Molecules  =  Moles × 6.022 × 10²³ Molecules.mol⁻¹

Putting value of moles,

     Number of C₄H₁₀ Molecules  =  0.0258 mol × 6.022 × 10²³ Molecules.mol⁻¹

                 Number of C₄H₁₀ Molecules  =  1.55 × 10²² C₄H₁₀ Molecules

Step 3: Calculate Number of Carbon Atoms:

As,

                            1 Molecule of C₄H₁₀ contains  =  4 Atoms of Carbon

So,

          1.55 × 10²² C₄H₁₀ Molecules will contain  =  X Atoms of Carbon

Solving for X,

 X =  (1.55 × 10²² C₄H₁₀ Molecules × 4 Atoms of Carbon) ÷ 1 Molecule of C₄H₁₀

X  =  6.21 × 10²² Atoms of Carbon

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