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3 years ago

Where does a wolf get energy from?

Chemistry

2 answers:

Setler79 [48]3 years ago

4 0

Answer:

Check Explanation

Explanation:

In this case the Mushroom would get energy from decomposing dead plants and animals in the soil/ground

The plant would get energy from the sun (simplified)

the rabbit would get energy from eating the plant

in this case the wolf would energy from eating the rabbit but in a normal setting the wolf could get energy from eating any animal

Rzqust [24]3 years ago

3 0

Answer:

Explanation: A wolf gets its energy from food and water

A rabbit gets its energy from food and water too

A plant get it energy from sunlight, rain

A mushroom also get its energy from dead plant/animal’s

If this wasn’t the answer you wanted I’m sorry

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A reaction will be spontaneous only at low temperatures if both ΔH and ΔS are negative. For a reaction in which ΔH = −320.1 kJ/m

Ganezh [65]

Answer:

This reaction is spontaneous for temperature lower than 3722.1 Kelvin or 3448.95 °C

Explanation:

<u>Step 1: </u>Data given

ΔH = −320.1 kJ/mol

ΔS = −86.00 J/K · mol.

<u>Step 2:</u> Calculate the temperature

ΔG<0 = spontaneous

ΔG= ΔH - TΔS

ΔH - TΔS <0

-320100 - T*(-86) <0

-320100 +86T < 0

-320100 < -86T

320100/86 > T

3722.1 > T

The temperature should be lower than 3722.1 Kelvin (= 3448.9 °C)

We can prove this with Temperature T = 3730 K

-320100 -3730*(-86) <0

-320100 + 320780 = 680 this is greater than 0 so it's non spontaneous

T = 3700 K

-320100 -3700*(-86) <0

-320100 + 318200 = -1900 this is lower than 0 so it's spontaneous

The temperature is quite high because of the big difference between ΔH and ΔS.

This reaction is spontaneous for temperature lower than 3722.1 Kelvin or 3448.95 °C

3 0

3 years ago

9 moles of Ammonia (NH3) are added to 50 L of H2O at a temperature of 29°C. The vapor pressure of water alone is 29.96 mmHg at 2

Westkost [7]

Answer: The vapor pressure of the solution at $29^0C$ is 29.86 mm Hg

Explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

$\frac{p^o-p_s}{p^o}=i\times x_2$

where,

$\frac{p^o-p_s}{p^o}$ = relative lowering in vapor pressure

i = Van'T Hoff factor = 1 (for non electrolytes)

$x_2$ = mole fraction of solute

= $\frac{\text {moles of solute}}{\text {total moles}}$

Given : 9 moles of $NH_3$ are dissolved in 50 L or 50000 ml of water

mass of water = $density\times volume = 1g/ml\times 50000ml=50000g$

moles of solvent (water) = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{50000g}{18g/mol}=2778moles$

Total moles = moles of solute + moles of solvent = 9 mol + 2778 mol = 2787

$x_2$ = mole fraction of solute

= $\frac{9}{2787}=3.2\times 10^{-3}$

$\frac{29.96-p_s}{29.96}=1\times 3.2\times 10^{-3}$

$p_s=29.86mmHg$

Thus the vapor pressure of the solution at $29^0C$ is 29.86 mm Hg

5 0

3 years ago

Which compound has the greater percentage of sodium: NaCl or NaBr

dimulka [17.4K]

NaCI has a greater % of sodium because the Chloride ion is closer in size to Na ion than Bromide ion is to Na ion. Therefore the NaCl lattice is more neatly and closely packed, leading to higher melting point.

6 0

3 years ago

List three evidences of physical and chemical change. ( If you know only one or two please write them anyways please)

LenaWriter [7]

Physical: ripping paper, braking a stick, cake batter mixing

Chemical: burning paper, spoiled milk, a penny turning green

3 0

4 years ago

Calculate the Empirical Formula for the following compound:<br> 0.300 mol of S and 0.900 mole of O.

ziro4ka [17]

Answer:

$\boxed {\boxed {\sf SO_3}}$

Explanation:

An empirical formula shows the smallest whole-number ratio of the atoms in a compound.

So, we must calculate this ratio. Since we are given the amounts of the elements in moles, we can do this in just 2 steps.

<h3>1. Divide </h3>

The first step is division. We divide the amount of moles for both elements by the <u>smallest amount of moles</u>.

There are 0.300 moles of sulfur and 0.900 moles of oxygen. 0.300 is smaller, so we divide both amounts by 0.300

Sulfur: 0.300/0.300= 1
Oxygen: 0.900/0.300= 3

<h3>2. Write Empirical Formula</h3>

The next step is writing the formula. We use the numbers we just found as the subscripts. These numbers go after the element's symbol in the formula. Remember sulfur is S and there is 1 mole and oxygen is O and there are 3 moles.

S₁O₃

This formula is technically correct, but we typically remove subscripts of 1 because no subscript implies 1 representative unit.

SO₃

$\bold {The \ empirical \ formula \ for \ the \ compound \ is \ SO_3}}$

6 0

3 years ago