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IgorLugansk [536]
3 years ago
15

The ability of organisms to function within a specific range of temperatures, salinities, or rainfall amounts?

Chemistry
1 answer:
balandron [24]3 years ago
3 0

Answer:

The range of variation under which a species can function and reproduce is called its tolerance range.

Explanation:

PLZZZ MARK ME BRAINLIESTTTTT

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THE GEOCENTRIC MODEL
Mars2501 [29]

Answer:

C

Explanation:

6 0
2 years ago
A 12 gram piece of metal is heated to 300 °C from 100 °C with 1120 Joules of energy. What is the specific heat of the metal?
Sergeeva-Olga [200]

Answer:

The specific heat for the metal is 0.466 J/g°C.

Explanation:

Given,

Q = 1120 Joules

mass = 12 grams

T₁ = 100°C

T₂ = 300°C

The specific heat for the metal can be calculated by using the formula

Q = (mass) (ΔT) (Cp)

ΔT = T₂ - T₁ = 300°C  - 100°C   = 200°C

Substituting values,

1120 = (12)(200)(Cp)

Cp = 0.466 J/g°C.

Therefore, specific heat of the metal is 0.466 J/g°C.

7 0
3 years ago
Both 1,2−dihydronaphthalene and 1,4−dihydronaphthalene may be selectively hydrogenated to 1,2,3,4−tetrahydronaphthalene. One of
pochemuha

Answer:

1,4-dihydro = 113 kJ·mol⁻¹

1,2-dihydro = 101 kJ·mol⁻¹

Explanation:

In 1,4-dihydronaphthalene, the 2,3-double bond is isolated from the benzene ring.

In 1,2-dihydronaphthalene, the 3,4-double bond is conjugated with the benzene ring.

Thus, 1,2-dihydronaphthalene is partially stabilized by resonance interactions between the ring and the double bond (think, styrene).

1,2-Dihydronaphthalene is at a lower energy level because of this stabilization.

The heat of hydrogenation of 1,2-dihydronaphthalene is therefore less than that of the 1,4-isomer when each is hydrogenated to the common product, 1,2,3,4-tetrahydronaphthalene.

8 0
2 years ago
Помогите пожалуйста или с первым вопросом или со вторым. Буду очень благодарна, т.к. это мой зачет... :С
Bess [88]
Hsiaqoanwbwiso iDisks
5 0
3 years ago
Look at sample problem 19.10 in the 8th ed Silberberg book. Write the Ksp expression. Find the concentrations of the ions you ne
777dan777 [17]

Answer:

Explanation:

From the information given:

CaF_2 \to Ca^{2+} + 2F^-

Ksp = 3.2 \times 10^{-11}

no of moles of Ca^{2+} = 0.01 L × 0.0010 mol/L

no of moles of Ca^{2+} = 1 \times 10^{-5} \ mol

no of moles of F^- = 0.01 L × 0.00010 mol/L

no of moles of F^- = 1 \times 10^{-6}\ mol

Total volume = 0.02 L

[Ca^{2+}}] = \dfrac{1\times10^{-5} \ mol}{0.02 \ L} \\ \\  \\  \[[Ca^{2+}}] = 0.0005 \ mol/L

[F^{-}] = \dfrac{(1\times 10^{-6} \ mol)}{0.02 \ L}

[F^{-}] = 5 \times 10^{-5}  \ mol/L

Q = [Ca^{2+}][F^-]^2 \\ \\ Q = 0.0005 \times (5\times 10^{-5})^2 \\ \\ Q = 1.25 \times 10^{-12}

Since Q<ksp, then there will no be any precipitation of CaF2

3 0
3 years ago
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