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2 years ago

Why do sea turtles rely on instincts more than mammals do?

Chemistry

1 answer:

qwelly [4]2 years ago

6 0

Answer:

Mammals do not have instincts, so they must learn all their behaviors from their parents in order to survive.

Sea turtles must survive on their own, while mammals spend time with their parents and learn how to survive as they grow.

Sea turtles are less intelligent than mammals and, thus, cannot learn new behaviors from their parents.

Explanation:

You might be interested in

Calculate the empirical formula for a compound containing 67.6% Hg 10.8% S 21.6% O

pashok25 [27]

Answer:

HgSO₄

Explanation:

% => g => moles => ratio => reduce => empirical ratio

%Hg = 67.6% => 67.6g/201g/mol = 0.34mol

%S = 10.8% => 10.8g/32g/mol = 0.34mol

%O = 21.6% => 21.6g/16g/mol = 1.35mol

Hg:S:O => 0.34:0.34:1.35

Reduce to whole number ratio by dividing by the smaller mole value...

Hg:S:O => 0.34/.34:0.34/.34:1.35/.34 => Empirical Ratio = 1:1:4

∴ Empirical Formula is HgSO₄

3 0

3 years ago

Solve the quadratic equation 2x^2+13x=15 by method of completing the square

mafiozo [28]

Answer:

x = 1, -7.5

Explanation:

2x² + 13x = 15

Divide the left side of the equation by 2

2(x² + 6.5x) = 15

Divide 6.5 by 2 and square the quotient

6.5/2 = 3.25

3.25² = 10.5625

Add 10.5625 to the left side

2(x² + 6.5x + 10.5625) = 15

Since you have a 2 outside the parentheses, you will be adding 10.5625•2 to the right side.

10.5625 • 2 = 21.125

2(x² + 6.5x + 10.5625) = 36.125

To factor (x² + 6.5x + 10.5625), add b/2 to x

b/2 = 6.5/2 = 3.25

2(x + 3.25)² = 36.125

Divide by 2

(x + 3.25)² = 18.0625

Square root.

(x + 3.25) = √18.0625

x + 3.25 = ±4.25

Subtract 3.25.

x = 4.25 - 3.25 = 1

x = -4.25 - 3.25 = -7.5

x = 1, -7.5

4 0

3 years ago

One mole of ice at 0°C is added to two moles of water at 50°C under a constant external pressure of 1 atm. This process is carri

kotykmax [81]

Answer:

T2 = 29.79°C

Explanation:

Equliibrium signifies that heat loss = heat gained

Heat gained by Ice;

H = ML

Mass, M = Number of moles * Molar mass = 1 * 18 = 18g

l = 6.01 k J m o l = 334 J/g

C = 4.186 J/g

H = 18(334)

H = 6012

Heat lost by water

H = MCΔT

H = 18 * 4.186 * (50 - T2)

H = 3767.4 - 75.348T2

Since H = H, we have;

6012 = 3767.4 - 75.348T2

- 75.348T2 = 3767 - 6012

T2 = 2245 / 75.348

T2 = 29.79°C

8 0

3 years ago

Suppose you have just added 100 ml of a solution containing 0.5 mol of acetic acid per liter to 400 ml of 0.5 m naoh. what is th

Tpy6a [65]

$pH = 13.5$

Explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

$\text{HAc} + \text{OH}^{-} \to \text{Ac}^{-} + \text{H}_2\text{O}$

The mixture would contain

$0.4 \times 0.5 - 0.1 \times 0.5 = 0.15 \; \text{mol}$ of $\text{OH}^{-}$ and
$0.1 \times 0.5 = 0.05 \; \text{mol}$ of $\text{Ac}^{-}$

if $\text{Ac}^{-}$ undergoes no hydrolysis; the solution is of volume $0.1 + 0.4 = 0.5 \; \text{L}$ after the mixing. The two species would thus be of concentration $0.30 \; \text{mol} \cdot \text{L}^{-1}$ and $0.10 \; \text{mol} \cdot \text{L}^{-1}$ , respectively.

Construct a RICE table for the hydrolysis of $\text{Ac}^{-}$ under a basic aqueous environment (with a negligible hydronium concentration.)

$\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}$

The question supplied the acid dissociation constant $pK_a$ for acetic acid $\text{HAc}$ ; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the base dissociation constant $pK_b$ for its conjugate base, $\text{Ac}^{-}$ . The following relationship relates the two quantities:

$pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})$

... where the water self-ionization constant $pK_w \approx 14$ under standard conditions. Thus $pK_{b} (\text{Ac}^{-}) = 14 - 4.7 = 9.3$ . By the definition of $pK_b$ :